Graphing Functions The ski lifts and the mountain both have a slope. (credit: modification of work "colorado springs zoo tram" by woodleywonderworks/Flickr, CC BY 2.0) Learning Objectives After completing this module, you should be able to: Graph functions using intercepts. Compute slope. Graph functions using slope and y -intercept. Graph horizontal and vertical lines. Interpret graphs of functions. Model applications using slope and y -intercept. In this section, we will expand our knowledge of graphing by graphing linear functions. There are many real-world scenarios that can be represented by graphs of linear functions. Imagine a chairlift going up at a ski resort. The journey a skier takes travelling up the chairlift could be represented as a linear function with a positive slope. The journey a skier takes down the slopes could be represented by a linear function with a negative slope. Graphing Functions Using Intercepts Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points. At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x -axis and the y -axis. These points are called the intercepts of a line. Let us review the graphs of the lines in . The table below lists where each of these lines crosses the x - and y -axis. Do you see a pattern? For each line, the y -coordinate of the point where the line crosses the x -axis is zero. The point where the line crosses the x -axis has the form ( a , 0 ) and is called the x -intercept of the line. The x -intercept occurs when y is zero. In each line, the x -coordinate of the point where the line crosses the y -axis is zero. The point where the line crosses the y -axis has the form ( 0 , b ) and is called the y -intercept of the line. The y -intercept occurs when x is zero. Figure The line crosses the x -axis at: Ordered Pair for this Point The line crosses the y -axis at: Ordered Pair for This Point Figure (a) 3 ( 3 , 0 ) 6 ( 0 , 6 ) Figure (b) 4 ( 4 , 0 ) − 3 ( 0 , − 3 ) Figure (c) 5 ( 5 , 0 ) − 5 ( 0 , − 5 ) Figure (d) 0 ( 0 , 0 ) 0 ( 0 , 0 ) General Figure a ( a , 0 ) b ( 0 , b ) Finding x - and y -Intercepts Find the x -intercept and y -intercept on the (a) and (b) graphs in . In , the graph crosses the x -axis at the point ( 4 , 0 ) . The x -intercept is ( 4 , 0 ) . The graph crosses the y -axis at the point ( 0 , 2 ) . The y -intercept is ( 0 , 2 ) . In , the graph crosses the x -axis at the point ( 2 , 0 ) . The x -intercept is ( 2 , 0 ) . The graph crosses the y -axis at the point ( 0 , − 6 ) . The y -intercept is ( 0 , − 6 ) . Graphing a Function Using Intercepts Find the intercepts of 2 x + y = 8 . Then graph the function using the intercepts. Let y = 0 to find the x -intercept, and let x = 0 to find the y -intercept. 2 x + y = 8 2 x + y = 8 To find the x -intercept, let y = 0 . 2 x + 0 = 8 To find the y -intercept, let x = 0 . 2 ( 0 ) + y = 8 Simplify. 2 x = 8 x = 4 Simplify. y = 8 The x -intercept is: ( 4 , 0 ) The y -intercept is: ( 0 , 8 ) Plot the intercepts to get the graph in . Computing Slope When graphing linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter. In mathematics, the measure of the steepness of a line is called the slope of the line. To find the slope of a line, we locate two points on the line whose coordinates are integers. Then we sketch a right triangle where the two points are vertices of the triangle and one side is horizontal and one side is vertical. Next, we measure or calculate the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run . We can assign a numerical value to the slope of a line by finding the ratio of the rise and run. The rise is the amount the vertical distance changes while the run measures the horizontal change, as shown in this illustration. Slope ( ) is a rate of change. To calculate slope ( m ) , use the formula m = rise run , where the rise measures the vertical change and the run measures the horizontal change. The concept of slope has many applications in the real world. In construction, the pitch of a roof, the slant of plumbing pipes, and the steepness of stairs are all applications of slope. As you ski or jog down a hill, you definitely experience slope. Finding the Slope from a Graph Find the slope of the line shown in . Step 1: Locate two points on the graph whose coordinates are integers, such as ( 0 , 5 ) and ( 3 , 3 ) . Starting at ( 0 , 5 ) , sketch a right triangle to ( 3 , 3 ) as shown in . Step 2: Count the rise; since it goes down, it is negative. The rise is −2. Step 3: Count the run. The run is 3. Step 4: Use the slope formula m = rise run substitute the values of the rise and run. m = − 2 3 The slope of the line is − 2 3 . The solution is y decreases by 2 units as x increases by 3 units. Sometimes we will need to find the slope of a line between two points when we don’t have a graph to measure the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing. First, we need to introduce some algebraic notation. We have seen that an ordered pair ( x , y ) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol ( x , y ) be used to represent two different points? Mathematicians use subscripts to distinguish such points. For example, ( x 1 , y 1 ) would be said aloud as “ x sub 1, y sub 1” and ( x 2 , y 2 ) read “ x sub 2, y sub 2.” The “sub” is a short way of saying “subscript.” We will use ( x 1 , y 1 ) to identify the first point and ( x 2 , y 2 ) to identify the second point in our slope equation. If we had more than two points, (if we were finding more than one slope), we could use ( x 3 , y 3 ), ( x 4 , y 4 ), and so on. Let’s review how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points ( 2 , 3 ) and ( 7 , 6 ) , as shown in . On the graph, we count the rise of 3 and the run of 5. Notice on the graph that that ( x 1 , y 1 ) is the point ( 2 , 3 ) and ( x 2 , y 2 ) is the point ( 7 , 6 ) . The rise can be found by subtracting the y -coordinates, 6 and 3, and the run can be found by subtracting the x -coordinates 7 and 2. m = rise run = 6 − 3 7 − 2 = 3 5 We have shown that m = y 2 − y 1 x 2 − x 1 is really another version of m = rise run . We can use this formula to find the slope of a line. To find the slope of the line between two points ( x 1 , y 1 ) and ( x 2 , y 2 ), use the formula m = y 2 − y 1 x 2 − x 1 Finding the Slope of the Line Using Points Use the slope formula to find the slope of the line through the points (−2, −3) and (−7, 4). We’ll call (−2, −3) point 1 and (−7, 4) point 2. Step 1: Use the slope formula: m = y 2 − y 1 x 2 − x 1 Step 2: Substitute the values: m = 4 − ( − 3 ) − 7 − ( − 2 ) Step 3: Simplify: m = 7 − 5 = − 7 5 Step 4: Verify the slope on the graph shown in . Graphing Functions Using Slope and y -Intercept We have graphed linear equations by plotting points and using intercepts. Once we see how an equation in slope-intercept form and its graph are related, we will have one more method we can use to graph lines. Review the graph of the equation y = 1 2 x + 3 in and find its slope and y -intercept. Graph of the equation y = 1 2 x + 3 . The vertical and horizontal lines in the graph show us the rise is 1 and the run is 2, respectively. Substituting into the slope formula: m = rise run = 1 2 The y -intercept is ( 0 , 3 ) . Look at the equation of this line. y = 1 2 x + 3 Look at the slope and y -intercept. slope m = 1 2 and y - intercept ( 0 , 3 ) When a linear equation is solved for y , the coefficient of the x term is the slope and the constant term is the y -coordinate of the y -intercept. We say that the equation y = 1 2 x + 3 is in slope-intercept form . Sometimes the slope-intercept form is called the y -form. Finding the Slope and y -Intercept of a Line Identify the slope and y -intercept of the line from the equation: y = − 4 7 x − 2 x + 3 y = 9 We compare our equation to the slope-intercept form of the equation. Step 1: Write the slope-intercept form of the equation of the line. y = m x + b Step 2: Write the equation of the line. y = − 4 7 x − 2 Step 3: Identify the slope. m = − 4 7 Step 4: Identify the y -intercept. y - intercept is ( 0 , − 2 ) When an equation of a line is not given in slope-intercept form, our first step will be to solve the equation for y . Step 1: Solve for y . x + 3 y = 9 Step 2: Subtract x from each side. 3 y = − x + 9 Step 3: Divide both sides by 3. 3 y 3 = − x + 9 3 Step 4: Simplify. y = − 1 3 x + 3 Step 5: Write the slope-intercept form of the equation of the line. y = m x + b Step 6: Write the equation of the line. y = − 1 3 x + 3 Step 7: Identify the slope. m = − 1 3 Step 8: Identify the y -intercept. y - intercept is ( 0 , 3 ) . Graphing the Slope and y -Intercept Graph the line of the equation y = − x + 4 using its slope and y -intercept. The equation is in slope-intercept form y = m x + b . y = − x + 4 . Step 1: Identify the slope and y -intercept. m = − 1 , y -intercept is ( 0 , 4 ) . Step 2: Plot the y -intercept on the coordinate system ( ). Identify the rise over the run. m = − 1 = − 1 1 Count out the rise and run to mark the second point. rise − 1 , run 1 Graphing Horizontal and Vertical Lines Some linear equations have only one variable. They may have just x without the y , or just y without an x . This changes how we make a table of values to get the points to plot. Let us consider the equation x = − 3 . This equation has only one variable, x . The equation says that x is always equal to − 3 , so its value does not depend on y . No matter what the value of y is, the value of x is always − 3 . To make a table of values, write − 3 in for all the x -values. Then choose any values for y . Since x does not depend on y , you can choose any numbers you like. But to fit the points on our coordinate graph, we will use 1, 2, and 3 for the y -coordinates in the table below. x = − 3 x y ( x , y ) −3 1 ( − 3 , 1 ) − 3 2 ( − 3 , 2 ) − 3 3 ( − 3 , 3 ) Plot the points from the table and connect them with a straight line ( ). Notice that we have graphed a vertical line. Graph of x = − 3 What is the slope? If we take the two points ( − 3 , 3 ) and ( − 3 , 1 ) then the rise is 2 and the run is 0. Using the slope formula we get: m = rise run = 2 0 The slope is undefined since division by zero is undefined. We say that the slope of the vertical line x = − 3 is undefined. The slope of any vertical line x = a (where a is any number) will be undefined. What if the equation has y but no x ? Let’s graph the equation y = 4 . This time the y -value is a constant, so in this equation, y does not depend on x . Fill in 4 for all the y values in the table below and then choose any values for x . We will use 0, 2, and 4 for the x -coordinates. y = 4 x y ( x , y ) 0 4 ( 0 , 4 ) 2 4 ( 2 , 4 ) 4 4 ( 4 , 4 ) In , we have graphed a horizontal line passing through the y -axis at 4. Graph of y = 4 What is the slope? If we take the two points ( 2 , 4 ) and ( 4 , 4 ) then the rise is 0 and the run is 2. Using the slope formula, we get m = rise run = 0 2 = 0 . The slope of the horizontal line y = 4 is 0. The slope of any horizontal line y = b (where b is any number) will be 0. When the y -coordinates are the same, the rise is 0. Graphing A Vertical Line Graph: x = 2 . The equation has only one variable, x , and x is always equal to 2. We create a table where x is always 2 and then put in any values for y . The graph is a vertical line passing through the x -axis at 2 ( ). x = 2 x y ( x , y ) 2 1 ( 2 , 1 ) 2 2 ( 2 , 2 ) 2 3 ( 2 , 3 ) Graphing A Horizontal Line Graph: y = − 1 . The equation y = − 1 has only one variable, y . The value of y is constant. All the ordered pairs in the next table have the same y -coordinate. The graph is a horizontal line passing through the y -axis at −1 ( ). y = − 1 x y ( x , y ) 0 − 1 ( 0 , − 1 ) 3 − 1 ( 3 , − 1 ) − 3 − 1 ( − 3 , − 1 ) The table below summarizes all the methods we have used to graph lines. Interpreting Graphs of Functions An important yet often overlooked area in algebra involves interpreting graphs. Oftentimes in math classes, students are given mathematical functions and can make graphs to represent them. But the interpretation of graphs is a more applicable skill to the real world. Being able to “read” a graph—understanding its domain and range, what the intercepts mean, and what the slope (or curve) means— that's a real-world skill. Interpreting a Graph In the x -axis on the graph represents the 120-minute bike ride Juan went on. The y -axis represents how far away he was from his home. Interpret the x - and y -intercept. For each segment, find the slope. Create an interpretation of this graph (i.e., make up a story that goes with it). ( 0 , 0 ) is the x - and y -intercept and represents Juan at home before his bike ride. The distance from home is 0 miles and 0 minutes have passed. In the first 30 minutes, the slope is 1 5 and indicates Juan is traveling 1 mile for every 5 minutes. Between 30 and 60 minutes, the slope is 0 and indicates that he’s not riding the bike (the distance is not increasing). Then between 60 and 90 minutes, the slope is 1 5 again. Finally, after 90 minutes the slope is − 4 15 , meaning Juan is getting 4 miles closer to home every 15 minutes. Answers will vary. Juan left his house for a bike ride. After 30 minutes, he was 6 miles from home and he stopped for ice cream at his local ice-cream truck. He enjoyed his ice cream for 30 minutes. He then jumped back on his bike and rode to his friend’s house. He arrived there 30 minutes later. His friend’s house was 12 miles from his home. His friend was not home so he immediately turned around and quickly rode home in 45 minutes. Modeling Applications Using Slope and y -Intercept Many real-world applications are modeled by linear equations. We will review a few applications here so you can understand how equations written in slope-intercept form relate to real-world situations. Usually when a linear equation model uses real-world data, different letters are used for the variables instead of using only x and y . The variable names often remind us of what quantities are being measured. Also, we often need to extend the axes in our rectangular coordinate system to bigger positive and negative numbers to accommodate the data in the application. Converting Temperature The equation F = 9 5 C + 32 is used to convert temperatures from degrees Celsius ( C ) to degrees Fahrenheit ( F ). Find the Fahrenheit temperature for a Celsius temperature of 0°. Find the Fahrenheit temperature for a Celsius temperature of 20°. Interpret the slope and F -intercept of the equation. Graph the equation. Find the Fahrenheit temperature for a Celsius temperature of 0°. Find F when C = 0 . F = 9 5 ( 0 ) + 32 Simplify. F = 32 Find the Fahrenheit temperature for a Celsius temperature of 20°. Find F when C = 20 . F = 9 5 ( 20 ) + 32 Simplify. F = 36 + 32 Simplify. F = 68 Interpret the slope and F -intercept of the equation. Even though this equation uses F and C , it is still in slope-intercept form. y = m x + b F = m C + b F = 9 5 C + 32 The slope 9 5 means that the temperature Fahrenheit ( F ) increases 9 degrees when the temperature Celsius ( C ) increases 5 degrees. The F -intercept means that when the temperature is 0° on the Celsius scale, it is 32° on the Fahrenheit scale. Graph the equation. We will need to use a larger scale than our usual. Start at the F -intercept ( 0 , 32 ) , and then count out the rise of 9 and the run of 5 to get a second point as shown in . Calculating Driving Costs Sam drives a delivery van. The equation C = 0.5 d + 60 models the relation between his weekly cost, C , in dollars and the number of miles, d , that he drives. Find Sam’s cost for a week when he drives 0 miles. Find the cost for a week when he drives 250 miles. Interpret the slope and C -intercept of the equation. Graph the equation. Find Sam’s cost for a week when he drives 0 miles. Find C when d = 0. C = 0.5 ( 0 ) + 60 Simplify. C = 60 Sam’s costs are $60 when he drives 0 miles. Find the cost for a week when he drives 250 miles. Find C when d = 250 . C = 0.5 ( 250 ) + 60 Simplify. C = 185 Sam’s costs are $185 when he drives 250 miles. Interpret the slope and C -intercept of the equation. y = m x + b C = 0.5 d + 60 The slope, 0.5, means that the weekly cost, C , increases by $0.50 when the number of miles driven, d , increases by 1. The C -intercept means that when the number of miles driven is 0, the weekly cost is $60. Graph the equation ( ). We’ll need to use a larger scale than usual. Start at the C -intercept (0, 60). To count out the slope m = 0.5 , we rewrite it as an equivalent fraction that will make our graphing easier. m = 0.5 = 5 10 = 50 100 So to graph the next point go up 50 from the intercept of 60 and then to the right 100. The second point will be (100, 110). Check Your Understanding Key Terms intercepts of a line slope slope-intercept form Key Concepts Every linear function can be graphically represented by a unique line that shows all the solutions of the equation. The points where the graph of a line intersects the x -axis and y -axis are called the intercepts of the line. Most lines will have one x -intercept and one y -intercept. Only if the line is straight vertical (no y -intercept) or straight horizontal (no x -intercept) will it not have both intercepts. Note that a line that is straight vertical is not a function, but a line that is straight horizontal is a function. Since any two points determine a straight line, any linear function can be graphed if both intercepts are known. The slope of a linear function is the ratio of the vertical change divided by the horizontal change. It is often referred to as r i s e r u n . A formula for finding the slope of linear functions is y 2 − y 1 x 2 − x 1 , for any two points of the linear function ( x 1 , y 1 ) and ( x 2 , y 2 ) . Formulas To calculate slope ( m ) , use the formula m = r i s e r u n , where the rise measures the vertical change and the run measures the horizontal change. To find the slope of the line between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) , use the formula m = y 2 − y 1 x 2 − x 1

Math


Systems of Linear Equations in Two Variables Fruits and vegetables at a farmer’s market. (credit: “California Ave. Farmers’ Market” by Jun Seita/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Determine and show whether an ordered pair is a solution to a system of equations. Solve systems of linear equations using graphical methods. Solve systems of linear equations using substitution. Solve systems of linear equations using elimination. Identify systems with no solution or infinitely many solutions. Solve applications of systems of linear equations. In this section, we will learn how to solve systems of linear equations in two variables. There are several real-world scenarios that can be represented by systems of linear equalities. Suppose two friends, Andrea and Bart, go shopping at a farmers market to buy some vegetables. Andrea buys 2 tomatoes and 4 cucumbers and spends $2.00. Bart buys 4 tomatoes and 5 cucumbers and spends $2.95. What is the price of each vegetable? Determining If an Ordered Pair Is a Solution to a System of Equations When we solved linear equations in Linear Equations in One Variable with Applications and Linear Inequalities in One Variable with Applications , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations . In this section, we will focus our work on systems of two linear equations in two unknowns (variables) and applications of systems of linear equations. An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. { 2 x + y = 7 x − 2 y = 6 A linear equation in two variables, such as 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line. To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) that make both equations true. These are called the solutions of a system of equations . To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. Determining Whether an Ordered Pair Is a Solution to the System Determine whether the ordered pair is a solution to the system. { x − y = − 1 2 x − y = − 5 ( − 2 , − 1 ) ( − 4 , − 3 ) { x − y = − 1 2 x − y = − 5 We substitute x = − 2 and y = − 1 into both equations. x − y = − 1 − 2 − ( − 1 ) = ? − 1 − 1 = − 1 ✓ 2 x − y = − 5 2 ( − 2 ) - ( − 1 ) = ? − 5 - 3 ≠ − 5 ( − 2 , − 1 ) does not make both equations true. ( − 2 , − 1 ) is not a solution. We substitute x = − 4 and y = − 3 into both equations. x − y = − 1 − 4 − ( − 3 ) = ? − 1 − 1 = − 1 ✓ 2 x − y = − 5 2 • ( − 4 ) − ( − 3 ) = ? − 5 − 5 = − 5 ✓ ( − 4 , − 3 ) makes both equations true. ( − 4 , − 3 ) is a solution. Determining Whether an Ordered Pair Is a Solution to the System Determine whether the ordered pair is a solution to the system { y = 3 2 x + 1 2 x − 3 y = 7 ( − 4 , − 5 ) ( − 4 , 5 ) { y = 3 2 x + 1 2 x − 3 y = 7 Substitute − 4 for x and − 5 for y into both equations. − 5 = ? 3 2 ( − 4 ) + 1 − 5 = ? 3 ( − 2 ) + 1 − 5 = ? − 6 + 1 − 5 = − 5 ✓ 2 ( − 4 ) − 3 ( − 5 ) = ? 7 ( − 8 ) − ( − 15 ) = ? 7 − 8 + 15 = ? 7 7 = 7 ✓ ( − 4 , − 5 ) is a solution. { y = 3 2 x + 1 2 x − 3 y = 7 Substitute − 4 for x and 5 for y into both equations. 5 = ? 3 2 ( − 4 ) + 1 5 = ? 3 ( − 2 ) + 1 5 = ? − 6 + 1 5 ≠ − 5 2 ( − 4 ) − 3 ( 5 ) = ? 7 ( − 8 ) − ( 15 ) = ? 7 − 8 − 15 = ? 7 − 23 ≠ 7 ( − 4 , 5 ) is not a solution. Solving Systems of Linear Equations Using Graphical Methods We will use three methods to solve a system of linear equations. The first method we will use is graphing. The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what points the lines have in common, we will find the solution to the system. Most linear equations in one variable have one solution; but for some equations called contradictions , there are no solutions, and for other equations called identities , all numbers are solutions. Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in . Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end you will decide which method was the most convenient way to solve this system. The steps to use to solve a system of linear equations by graphing are shown here. Step 1: Graph the first equation. Step 2: Graph the second equation on the same rectangular coordinate system. Step 3: Determine whether the lines intersect, are parallel, or are the same line. Step 4: Identify the solution to the system. If the lines intersect, identify the point of intersection. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions. Step 5: Check the solution in both equations. Solving a System of Linear Equations by Graphing Solve this system of linear equations by graphing. { 2 x + y = 7 x − 2 y = 6 Step 1: Graph the first equation. To graph the first line, write the equation in slope-intercept form. 2 x + y = 7 y = - 2 x + 7 m = - 2 b = 7 { 2 x + y = 7 x - 2 y = 6 Step 2: Graph the second equation on the same rectangular coordinate system. To graph the second line, use intercepts. x - 2 y = 6 ( 0 , - 3 ) ( 6 , 0 ) Step 3: Determine whether the lines intersect, are parallel, or are the same line. Look at the graph of the lines. The lines intersect. Step 4 and Step 5: Identify the solution to the system. If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions. Since the lines intersect, find the point of intersection. Check the point in both equations. The lines intersect at ( 4 , - 1 ) . 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 8 - 1 = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ The solution is ( 4 , - 1 ) . Solving Systems of Linear Equations Using Substitution We will now solve systems of linear equations by the substitution method. We will use the same system we used for graphing. { 2 x + y = 7 x − 2 y = 6 We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy. Then, we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those! After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true. This process is summarized here: Step 1: Solve one of the equations for either variable. Step 2: Substitute the expression from Step 1 into the other equation. Step 3: Solve the resulting equation. Step 4: Substitute the solution in Step 3 into either of the original equations to find the other variable. Step 5: Write the solution as an ordered pair. Step 6: Check that the ordered pair is a solution to both original equations. Solving a System of Linear Equations Using Substitution Solve this system of linear equations by substitution: { 2 x + y = 7 x − 2 y = 6 Step 1: Solve one of the equations for either variable. We’ll solve the first equation for y . { 2 x + y = 7 x - 2 y = 6 2 x + y = 7 y = 7 − 2 x Step 2: Substitute the expression from Step 1 into the other equation. We replace y in the second equation with the expression 7 - 2 x . x - 2 y = 6 x - 2 ( 7 - 2 x ) = 6 Step 3: Solve the resulting equation. Now we have an equation with just 1 variable. We know how to solve this! x - 2 ( 7 - 2 x ) = 6 x - 14 + 4 x = 6 5 x = 20 x = 4 Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable. We’ll use the first equation and replace x with 4. 2 x + y = 7 2 ( 4 ) + y = 7 8 + y = 7 y = - 1 Step 5: Write the solution as an ordered pair. The ordered pair is ( x , y ). ( 4 , - 1 ) Step 6: Check that the ordered pair is a solution to both original equations. Substitute x = 4 , y = - 1 into both equations and make sure they are both true. 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ Both equations are true. ( 4 , - 1 ) is the solution to the system. Solving Systems of Linear Equations Using Elimination We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small, and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression. The third method of solving systems of linear equations is called the elimination method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there. The elimination method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal. For any expressions a , b , c , and d : if a = b and c = d then a + c = b + d . To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable. Notice how that works when we add these two equations together: { 3 x + y = 5 2 x − y = 0 _ 5 x = 5 The y ’s add to zero and we have one equation with one variable. Let us try another one: { x + 4 y = − 2 2 x + 5 y = − 2 This time we do not see a variable that can be immediately eliminated if we add the equations. But if we multiply the first equation by − 2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by − 2 . { − 2 ( x + 4 y ) = − 2 ( –2 ) 2 x + 5 y = − 2 Then rewrite the system of equations. { − 2 x − 8 y = 4 2 x + 5 y = − 2 Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations. { − 2 x − 8 y = 4 2 x + 5 y = − 2 _ − 3 y = 2 Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations. Here’s a summary of using the elimination method: Step 1: Write both equations in standard form. If any coefficients are fractions, clear them. Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites. Step 3: Add the equations resulting from Step 2 to eliminate one variable. Step 4: Solve for the remaining variable. Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Step 6: Write the solution as an ordered pair. Step 7: Check that the ordered pair is a solution to both original equations. Solving a System of Linear Equations Using Elimination Solve this system of linear equations by elimination: { 2 x + y = 7 x − 2 y = 6 Step 1: Write both equations in standard form. If any coefficients are fractions, clear them. Both equations are in standard form, A x + B y = C . There are no fractions. { 2 x + y = 7 x - 2 y = 6 Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites. We can eliminate the y ’s by multiplying the first equation by 2. Multiply both sides of 2 x + y = 7 by 2. { 2 x + y = 7 x - 2 y = 6 { 2 ( 2 x + y ) = 2 ( 7 ) x - 2 y = 6 Step 3: Add the equations resulting from Step 2 to eliminate one variable. We add the x ’s, y ’s, and constants. { 4 x + 2 y = 14 x - 2 y = 6 _ 5 x = 20 Step 4: Solve for the remaining variable. Solve for x . x = 4 Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Substitute x = 4 into the second equation, x - 2 y = 6 . Then solve for y . x - 2 y = 6 4 - 2 y = 6 - 2 y = 2 y = - 1 Step 6: Write the solution as an ordered pair. Write it as ( x , y ) . ( 4 , - 1 ) Step 7: Check that the ordered pair is a solution to both original equations. Substitute x = 4 , y = - 1 into 2 x + y = 7 and x - 2 y = 6 . Do they make both equations true? Yes! 2 x + y = 7 2 ( 4 ) + ( - 1 ) = ? 7 7 = 7 ✓ x - 2 y = 6 4 - 2 ( - 1 ) = ? 6 6 = 6 ✓ The solution is ( 4 , - 1 ) . Identifying Systems with No Solution or Infinitely Many Solutions In all the systems of linear equations so far, the lines intersected, and the solution was one point. In and , we will look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions. Solving a System of Linear Equations with No Solution Solve the system by a method of your choice: { y = 1 2 x − 3 x − 2 y = 4 Let us solve the system of linear equations by graphing. { y = 1 2 x − 3 x − 2 y = 4 To graph the first equation, we will use its slope and y -intercept. y = 1 2 x − 3 m = 1 2 b = − 3 To graph the second equation, we will use the intercepts. x − 2 y = 4 x y 0 −2 4 0 Graph the lines ( ). Determine the points of intersection. The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system. Solving a System of Linear Equations with Infinite Solutions Solve the system by a method of your choice: { y = 2 x − 3 − 6 x + 3 y = − 9 Let us solve the system of linear equations by graphing. { y = 2 x − 3 − 6 x + 3 y = − 9 Find the slope and y -intercept of the first equation. y = 2 x − 3 m = 2 b = − 3 Find the intercepts of the second equation. − 6 x + 3 y = − 9 x y 0 −3 3 2 0 Graph the lines ( ). The lines are the same! Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to this system. In the previous example, if you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system. We say the two lines are coincident. Coincident lines have the same slope and same y -intercept. A system of equations that has at least one solution is called a consistent system . A system with parallel lines has no solution. We call a system of equations like this an inconsistent system . It has no solution. We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent. If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines. Let us sum this up by looking at the graphs of the three types of systems. See and the table that follows Lines Intersecting Parallel Coincident Number of Solutions 1 point No solution Infinitely many Consistent/Inconsistent Consistent Inconsistent Consistent Dependent/Independent Independent Independent Dependent Using Matrices and Cramer’s Rule to Solve Systems of Linear Equations An m by n matrix is an array with m rows and n columns, where each item in the matrix is a number. Matrices are used for many things, but one thing they can be used for is to represent systems of linear equations. For example, the system of linear equations { 2 x + y = 7 x − 2 y = 6 can be represented by the following matrix: [ 2 1 7 1 − 2 6 ] To use Cramer’s Rule, you need to be able to take the determinant of a matrix. The determinant of a 2 by 2 matrix A , denoted | A | , is | A | = | a 11 a 12 a 21 a 22 | = ( a 11 × a 22 ) + ( a 21 × a 12 ) For example, the determinant of the matrix | 2 1 3 − 2 | = ( 2 × − 2 ) − ( 3 × 1 ) = − 4 − 3 = − 7. Cramer’s Rule involves taking three determinants: The determinant of the first two columns, denoted | D | ; The determinant of the first column and the third column, denoted | D y | ; The determinant of the third column and the first column, denoted | D x | . Going back to the original matrix [ 2 1 7 1 − 2 6 ] | D | = | 2 1 1 − 2 | = − 4 − 1 = − 5 | D y | = | 2 7 1 6 | = 12 − 7 = 5 | D x | = | 7 1 6 − 2 | = − 14 − 6 = − 20 Now Cramer’s Rule for the solution of the system will be: x = | D x | | D | , y = | D y | | D | Putting in the values for these determinants, we have x = − 20 − 5 = 4 ; y = 5 − 5 = − 1. The solution to the system is the ordered pair ( 4 , − 1 ) . Solving Applications of Systems of Linear Equations Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we will first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system. Step 1 : Read the problem. Make sure all the words and ideas are understood. Step 2: Identify what we are looking for. Step 3: Name what we are looking for. Choose variables to represent those quantities. Step 4: Translate into a system of equations. Step 5: Solve the system of equations using good algebra techniques. Step 6: Check the answer in the problem and make sure it makes sense. Step 7: Answer the question with a complete sentence. Applying System to a Real-World Application Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 a year plus $15 for each training session. Option B would pay her $10,000 a year plus $40 for each training session. How many training sessions would make the salary options equal? Step 1: Read the problem. Step 2: Identify what we are looking for. We are looking for the number of training sessions that would make the pay equal. Step 3: Name what we are looking for. Let s = Heather’s salary , and n = the number of training sessions Step 4: Translate into a system of equations. Option A would pay her $25,000 plus $15 for each training session. s = 25,000 + 15 n Option B would pay her $10,000 + $40 for each training session. s = 10,000 + 40 n The system is shown. { s = 25,000 + 15 n s = 10,000 + 40 n Step 5: Solve the system of equations. We will use substitution. Substitute 25,000 + 15 n for s in the second equation s = 25,000 + 15 n s = 10,000 + 40 n Solve for n . 25,000 + 15 n = 10,000 + 40 n 25,000 = 10,000 + 25 n 15,000 = 25 n 600 = n Step 6: Check the answer. Are 600 training sessions a year reasonable? Are the two options equal when n = 600 ? Substitute into each equation. s = 25 , 000 + 15 ( 600 ) = 34 , 000 s = 10 , 000 + 40 ( 600 ) = 34 , 000 Step 7: Answer the question. The salary options would be equal for 600 training sessions. Practice with Solving Applications of Systems of Equations Applications of Systems of Linear Equations Check Your Understanding Key Terms system of linear equations solutions of a system of equations contradictions identities coincident lines consistent system of linear equations inconsistent system of linear equations Key Concepts To solve a system of linear equations means finding the point or points where the two linear equations intersect. Two lines can intersect at one point, no points if they are parallel, or every point if they are the same equation. Systems of linear equations can be solved by graphing, by using substitution, or by using the elimination method. Videos Practice with Solving Applications of Systems of Equations Applications of Systems of Linear Equations

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Systems of Linear Inequalities in Two Variables Many college students find part-time jobs at places such as coffee shops to help pay for college. (credit: modification of work “TULLY’s COFFEE” by MIKI Yoshihito/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Demonstrate whether an ordered pair is a solution to a system of linear inequalities. Solve systems of linear inequalities using graphical methods. Graph systems of linear inequalities. Interpret and solve applications of linear inequalities. In this section, we will learn how to solve systems of linear inequalities in two variables. In Systems of Linear Equations in Two Variables , we learned how to solve for systems of linear equations in two variables and found a solution that would work in both equations. We can solve systems of inequalities by graphing each inequality (as discussed in Graphing Linear Equations and Inequalities ) and putting these on the same coordinate system. The double-shaded part will be our solution to the system. There are many real-life examples for solving systems of linear inequalities. Consider Ming who has two jobs to help her pay for college. She works at a local coffee shop for $7.50 per hour and at a research lab on campus for $12 per hour. Due to her busy class schedule, she cannot work more than 15 hours per week. If she needs to make at least $150 per week, can she work seven hours at the coffee shop and eight hours in the lab? Determining If an Ordered Pair Is a Solution of a System of Linear Inequalities The definition of a system of linear inequalities is similar to the definition of a system of linear equations. A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here. { x + 4 y ≥ 10 3 x − 2 y < 12 To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) that make both inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the xy -coordinate system that includes all the points whose ordered pairs make the inequalities true. To determine if an ordered pair is a solution to a system of two inequalities, substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system. Determining Whether an Ordered Pair Is a Solution to a System Determine whether the ordered pair is a solution to the system: { x + 4 y ≥ 10 3 x − 2 y < 12 ( − 2 , 4 ) ( 3 , 1 ) Is the ordered pair ( − 2 , 4 ) a solution? We substitute x = − 2 and y = 4 into both inequalities. x + 4 y ≥ 10 - 2 + 4 ( 4 ) ≥ ? 10 14 ≥ 10 true 3 x − 2 y < 12 3 ( - 2 ) − 2 ( 4 ) < ? 12 − 14 < 12 true The ordered pair ( − 2 , 4 ) made both inequalities true. Therefore ( − 2 , 4 ) is a solution to this system. Is the ordered pair ( 3 , 1 ) a solution? We substitute x = 3 and y = 1 into both inequalities. x + 4 y ≥ 10 3 + 4 ( 1 ) ≥ ? 10 7 ≥ 10 false 3 x − 2 y < 12 3 ( 3 ) − 2 ( 1 ) < ? 12 7 < 12 true The ordered pair ( 3 , 1 ) made one inequality true, but the other one false. Therefore ( 3 , 1 ) is not a solution to this system. Solving Systems of Linear Inequalities Using Graphical Methods The solution to a single linear inequality was the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. We will review graphs of linear inequalities and solve the linear inequality from its graph. Solving a System of Linear Inequalities by Graphing Use to solve the system of linear inequalities: { y ≥ 2 x − 1 y < x + 1 To solve the system of linear inequalities we look at the graph and find the region that satisfies BOTH inequalities. To do this we pick a test point and check. Let's us pick ( − 1 , − 1 ) . Is ( - 1 , - 1 ) a solution to y ≥ 2 x − 1 ? - 1 ≥ ? 2 ( - 1 ) − 1 − 1 ≥ − 3 true Is ( - 1 , - 1 ) a solution to y < x + 1 ? - 1 < ? - 1 + 1 − 1 < 0 true The region containing ( − 1 , − 1 ) is the solution to the system of linear inequalities. Notice that the solution is all the points in the area shaded twice, which appears as the darkest shaded region. Graphing Systems of Linear Inequalities We learned that the solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region by graphing, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. Step 1: Graph the first inequality. Graph the boundary line. Shade in the side of the boundary line where the inequality is true. Step 2: On the same grid, graph the second inequality. Graph the boundary line. Shade in the side of that boundary line where the inequality is true. Step 3: The solution is the region where the shading overlaps. Step 4: Check by choosing a test point. Solving a System of Linear Inequalities by Graphing Solve the system by graphing: { x − y > 3 y < − 1 5 x + 4 Graph x − y > 3 by graphing x − y = 3 and testing a point ( ). The intercepts are x = 3 and y = − 3 and the boundary line will be dashed. Test ( 0 , 0 ) which makes the inequality false so shade the side that does not contain ( 0 , 0 ) . Graph y < − 1 5 x + 4 by graphing y = − 1 5 x + 4 using the slope m = − 1 5 and y -intercept b = 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Choose a test point in the solution and verify that it is a solution to both inequalities. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region. Graphing a System of Linear Inequalities Solve the system by graphing: { x − 2 y < 5 y > − 4 Graph x − 2 y < 5 by graphing x − 2 y = 5 ( ) and testing a point. The intercepts are x = 5 and y = − 2.5 and the boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Graph y > − 4 by graphing y = − 4 and recognizing that it is a horizontal line through y = − 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true so shade the side that contains ( 0 , 0 ) . The point ( 0 , 0 ) is in the solution, and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice, which appears as the darkest shaded region. Systems of linear inequalities where the boundary lines are parallel might have no solution. We will see this in the next example. Graphing Parallel Boundary Lines with No Solution Solve the system by graphing: { 4 x + 3 y ≥ 12 y < − 4 3 x + 1 Graph 4 x + 3 y ≥ 12 , by graphing 4 x + 3 y = 12 ( ) and testing a point. The intercepts are x = 3 and y = 4 and the boundary line will be solid. Test ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) . Graph y < − 4 3 x + 1 by graphing y = − 4 3 x + 1 using the slope m = − 4 3 and y -intercept b = 1 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . No shared point exists in both shaded regions, so the system has no solution. Some systems of linear inequalities where the boundary lines are parallel will have a solution. We will see this in the next example. Graphing Parallel Boundary Lines with a Solution Solve the system by graphing: { y > 1 2 x − 4 x − 2 y < − 4 Graph y > 1 2 x − 4 by graphing y = 1 2 x − 4 using the slope m = 1 2 and the y -intercept b = − 4 ( ). The boundary line will be dashed. Test ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) . Graph x − 2 y < − 4 by graphing x − 2 y = − 4 ( ) and testing a point. The intercepts are x = − 4 and y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalities. Test ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) . No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice which is also the solution to x − 2 y < − 4 . Interpreting and Solving Applications of Linear Inequalities When solving applications of systems of inequalities, first translate each condition into an inequality. Then graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so add inequalities to the system as additional requirements. Applying Linear Inequalities to Calculating Photo Costs A photographer sells their prints at a booth at a street fair. At the start of the day, they want to have at least 25 photos to display at their booth. Each small photo they display costs $4 and each large photo costs $10. They do not want to spend more than $200 on photos to display. Write a system of inequalities to model this situation. Graph the system. Could they display 10 small and 20 large photos? Could they display 20 large and 10 small photos? Let x = the number of small photos and y = the number of large photos . To find the system of equations translate the information. They want to have at least 25 photos. The number of small plus the number of large should be at least 25. x + y ≥ 25 $4 for each small and $10 for each large must be no more than $200 4 x + 10 y ≤ 200 The number of small photos must be greater than or equal to 0. x ≥ 0 The number of large photos must be greater than or equal to 0. y ≥ 0 We have our system of equations. { x + y ≥ 25 4 x + 10 y ≤ 200 x ≥ 0 y ≥ 0 Since x ≥ 0 and y ≥ 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only Quadrant I. To graph x + y ≥ 25 , graph x + y = 25 as a solid line. Choose ( 0 , 0 ) as a test point. Since it does not make the inequality true, shade the side that does not include the point ( 0 , 0 ) . To graph 4 x + 10 y ≤ 200 , graph 4 x + 10 y = 200 as a solid line. Choose ( 0 , 0 ) as a test point. Since it does make the inequality true, shade (bottom left) the side that include the point ( 0 , 0 ) . The solution of the system is the region of that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x -axis from ( 25 , 0 ) to ( 55 , 0 ) . To determine if 10 small and 20 large photos would work, we look at the graph to see if the point ( 10 , 20 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, so the photographer would not display 10 small and 20 large photos. To determine if 20 small and 10 large photos would work, we look at the graph to see if the point ( 20 , 10 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so the photographer could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality. Solving Systems of Linear Inequalities by Graphing Systems of Linear Inequalities Check Your Understanding Key Terms system of linear inequalities Key Concepts To solve a system of linear inequalities means to find the area(s) where the points in that area make all the linear inequalities true. Systems of linear inequalities can be solved by graphing the linear equations associated with the inequalities, then 'testing' points to see whether the values of the point make the equation true or not. Videos Solving Systems of Linear Inequalities by Graphing Systems of Linear Inequalities

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Linear Programming The aftermath of an earthquake and tsunami. (credit: modification of work "Earthquake and Tsunami Japan" by Climate and Ecosystems Change Adaptation Research University Network/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compose an objective function to be minimized or maximized. Compose inequalities representing a system application. Apply linear programming to solve application problems. Imagine you hear about some natural disaster striking a far-away country; it could be an earthquake, a fire, a tsunami, a tornado, a hurricane, or any other type of natural disaster. The survivors of this disaster need help—they especially need food, water, and medical supplies. You work for a company that has these supplies, and your company has decided to help by flying the needed supplies into the disaster area. They want to maximize the number of people they can help. However, there are practical constraints that need to be taken into consideration; the size of the airplanes, how much weight each airplane can carry, and so on. How do you solve this dilemma? This is where linear programming comes into play. Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints. As a field, linear programming began in the late 1930s and early 1940s. It was used by many countries during World War II; countries used linear programming to solve problems such as maximizing troop effectiveness, minimizing their own casualties, and maximizing the damage they could inflict upon the enemy. Later, businesses began to realize they could use the concept of linear programming to maximize output, minimize expenses, and so on. In short, linear programming is a method to solve problems that involve finding a maximum or minimum where a linear function is constrained by various factors. A Mathematician Invents a “Tsunami Cannon” On December 26, 2004, a massive earthquake occurred in the Indian Ocean. This earthquake, which scientists estimate had a magnitude of 9.0 or 9.1 on the Richter Scale, set off a wave of tsunamis across the Indian Ocean. The waves of the tsunami averaged over 30 feet (10 meters) high, and caused massive damage and loss of life across the coastal regions bordering the Indian Ocean. Usama Kadri works as an applied mathematician at Cardiff University in Wales. His areas of research include fluid dynamics and non-linear phenomena. Lately, he has been focusing his research on the early detection and easing of the effects of tsunamis. One of his theories involves deploying a series of devices along coastlines which would fire acoustic-gravity waves (AGWs) into an oncoming tsunami, which in theory would lessen the force of the tsunami. Of course, this is all in theory, but Kadri believes it will work. There are issues with creating such a device: they would take a tremendous amount of electricity to generate an AGW, for instance, but if it would save lives, it may well be worth it. Compose an Objective Function to Be Minimized or Maximized An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized. Composing an Objective Function for Selling Two Products Miriam starts her own business, where she knits and sells scarves and sweaters out of high-quality wool. She can make a profit of $8 per scarf and $10 per sweater. Write an objective function that describes her profit. Let x represent the number of scarves sold, and let y represent the number of sweaters sold. Let P represent profit. Since each scarf has a profit of $8 and each sweater has a profit of $10, the objective function is P = 8 x + 10 y . Composing an Objective Function for Production William’s factory produces two products, widgets and wadgets. It takes 24 minutes for his factory to make 1 widget, and 32 minutes for his factory to make 1 wadget. Write an objective function that describes the time it takes to make the products. Let x equal the number of widgets made; let y equal the number of wadgets made; let T represent total time. The objective function is T = 24 x + 32 y . Composing Inequalities Representing a System Application For our two examples of profit and production, in an ideal world the profit a person makes and/or the number of products a company produces would have no restrictions. After all, who wouldn’t want to have an unrestricted profit? However in reality this is not the case; there are usually several variables that can restrict how much profit a person can make or how many products a company can produce. These restrictions are called constraints . Many different variables can be constraints. When making or selling a product, the time available, the cost of manufacturing and the amount of raw materials are all constraints. In the opening scenario with the tsunami, the maximum weight on an airplane and the volume of cargo it can carry would be constraints. Constraints are expressed as linear inequalities; the list of constraints defined by the problem forms a system of linear inequalities that, along with the objective function, represent a system application. Representing the Constraints for Selling Two Products Two friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for profit and the two constraints. Let x represent the number of scarves sold, and let y represent the number of sweaters sold. There are two constraints: the number of items the business can make in a day (a maximum of 8) and the number of bags of knitting wool they can use per day (a maximum of 27). The first constraint (total number of items in a day) is written as: x + y ≤ 8 Since each scarf takes 3 bags of knitting wool and each sweater takes 4 bags of knitting wool, the second constraint, total bags of knitting wool per day, is written as: 3 x + 4 y ≤ 27 In summary, here are the equations that represent the new business: P = 8 x + 10 y ; This is the profit equation: The business makes $8 per scarf and $10 per sweater. x + y ≤ 8 3 x + 4 y ≤ 27 Representing Constraints for Production A factory produces two products, widgets and wadgets. It takes 24 minutes for the factory to make 1 widget, and 32 minutes for the factory to make 1 wadget. Research indicates that long-term demand for products from the factory will result in average sales of 12 widgets per day and 10 wadgets per day. Because of limitations on storage at the factory, no more than 20 widgets or 17 wadgets can be made each day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for time and the two constraints. Let x equal the number of widgets made; let y equal the number of wadgets made. Based on the long-term demand, we know the factory must produce a minimum of 12 widgets and 10 wadgets per day. We also know because of storage limitations, the factory cannot produce more than 20 widgets per day or 17 wadgets per day. Writing those as inequalities, we have: x ≥ 12 y ≥ 10 x ≤ 20 y ≤ 17 The number of widgets made per day must be between 12 and 20, and the number of wadgets made per day must be between 10 and 17. Therefore, we have: 12 ≤ x ≤ 20 10 ≤ y ≤ 17 The system is: T = 24 x + 32 y 12 ≤ x ≤ 20 10 ≤ y ≤ 17 T is the variable for time; it takes 24 minutes to make a widget and 32 minutes to make a wadget. Applying Linear Programming to Solve Application Problems There are four steps that need to be completed when solving a problem using linear programming. They are as follows: Step 1: Compose an objective function to be minimized or maximized. Step 2: Compose inequalities representing the constraints of the system. Step 3: Graph the system of inequalities representing the constraints. Step 4: Find the value of the objective function at each corner point of the graphed region. The first two steps you have already learned. Let’s continue to use the same examples to illustrate Steps 3 and 4. Solving a Linear Programming Problem for Two Products Three friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Determine the number of scarves and sweaters they should make each day to maximize their profit. Step 1: Compose an objective function to be minimized or maximized. From , the objective function is P = 8 x + 10 y . Step 2: Compose inequalities representing the constraints of the system. From , the constraints are x + y ≤ 8 and 3 x + 4 y ≤ 27 . Step 3: Graph the system of inequalities representing the constraints. Using methods discussed in Graphing Linear Equations and Inequalities , the graphs of the constraints are shown below. Because the number of scarves ( x ) and the number of sweaters ( y ) both must be non-negative numbers (i.e., x ≥ 0 and y ≥ 0 ), we need to graph the system of inequalities in Quadrant I only. shows each constraint graphed on its own axes, while shows the graph of the system of inequalities (the two constraints graphed together). In , the large shaded region represents the area where the two constraints intersect. If you are unsure how to graph these regions, refer back to Graphing Linear Equations and Inequalities . Graphs of each constraint Graph of the System of Inequalities Step 4: Find the value of the objective function at each corner point of the graphed region. The “graphed region” is the area where both of the regions intersect; in , it is the large shaded area. The “corner points” refer to each vertex of the shaded area. Why the corner points? Because the maximum and minimum of every objective function will occur at one (or more) of the corner points. shows the location and coordinates of each corner point. Graph of Region with Corner Points Three of the four points are readily found, as we used them to graph the regions; the fourth point, the intersection point of the two constraint lines, will have to be found using methods discussed in Systems of Linear Equations in Two Variables , either using substitution or elimination. As a reminder, set up the two equations of the constraint lines: 3 x + 4 y = 27 x + y = 8 For this example, substitution will be used. x + y = 8 y = 8 - x . Substituting 8 - x into the first equation for y , we have 3 x + 4 ( 8 − x ) = 27 3 x + 32 − 4 x = 27 − x = − 5 x = 5 Now, substituting the 5 in for x in either equation to solve for y . Choosing the second equation, we have: 5 + y = 8 y = 3 Therefore, x = 5 , and y = 3 . To find the value of the objective function, P = 8 x + 10 y , put the coordinates for each corner point into the equation and solve. The largest solution found when doing this will be the maximum value, and thus will be the answer to the question originally posed: determining the number of scarves and sweaters the new business should make each day to maximize their profit. Corner ( x , y ) Objective Function P = 8 x + 1 0 y ( 0 , 0 ) P = 8 ( 0 ) + 10 ( 0 ) = 0 ( 0 , 6.75 ) P = 8 ( 0 ) + 10 ( 6.75 ) = 67.5 ( 5 , 3 ) P = 8 ( 5 ) + 10 ( 3 ) = 40 + 30 = 70 ( 8 , 0 ) P = 8 ( 8 ) + 10 ( 0 ) = 64 The maximum value for the profit P occurs when x = 5 and y = 3 . This means that to maximize their profit, the new business should make 5 scarves and 3 sweaters every day. Leonid Kantorovich Leonid Vitalyevich Kantorovich was born January 19, 1912, in St. Petersburg, Russia. Two major events affected young Leonid’s life: when he was five, the Russian Revolution began, making life in St. Petersburg very difficult; so much so that Leonid’s family fled to Belarus for a year. When Leonid was 10, his father died, leaving his mother to raise five children on her own. Despite the hardships, Leonid showed incredible mathematical ability at a young age. When he was only 14, he enrolled in Leningrad State University to study mathematics. Four years later, at age 18, he graduated with what would be equivalent to a Ph.D. in mathematics. Although his primary interests were in pure mathematics, in 1938 he began working on problems in economics. Supposedly, he was approached by a local plywood manufacturer with the following question: how to come up with a work schedule for eight lathes to maximize output, given the five different kinds of plywood they had at the factory. By July 1939, Leonid had come up with a solution, not only to the lathe scheduling problem but to other areas as well, such as an optimal crop rotation schedule for farmers, minimizing waste material in manufacturing, and finding optimal routes for transporting goods. The technique he discovered to solve these problems eventually became known as linear programming. He continued to use this technique for solving many other problems involving optimization, which resulted in the book The Best Use of Economic Resources , which was published in 1959. His continued work in linear programming would ultimately result in him winning the Nobel Prize of Economics in 1975. Check Your Understanding Key Terms linear programming objective function constraint Key Concepts Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints. An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized. In linear programming, a constraint is a restriction that affects the maximum or minimum values of an objective function. Through the creation of objective functions and restraints, a linear system can be developed and solved through linear programming. Projects Ratio and Proportion—Comparing Prices, Part 1 Go to your favorite coffee shops and find out what a same sized drink costs at each. You can do something similar for pizza as well. Find the unit rate (i.e., price per ounce or price per square inch). For example, go to your favorite coffee place and find the price per units on all their large coffee drinks. Or go to your favorite pizza place and compare prices of all their extra-large pizzas (by price per square inch). Write a report on the best deals. Ratio and Proportion—Comparing Prices, Part 2 Rather than comparing prices of different, but same sized drinks (or pizzas), compare unit prices of the same drinks but of different sizes. Find out what the best bargain is based on price per ounce, price per square inch, etc. For example, compare the prices of your favorite soft drink sold at a local store, but in various sizes (i.e., 12-ounce can, 16-ounce bottle, 20-ounce bottle, 1-liter bottle, and multipacks). Or go to a pizza place and find out what the best bargain is on their menu, based on price per square inch of pizza. Write a report on the best deals. Systems of Linear Inequalities—Comparing Cell Phone Plans Go to the websites of different cell phone companies and compare their plans. Write a report on “the best deals. "Best Deals” doesn’t necessarily mean “cheapest.” You will need to look at what each company provides concerning restrictions (constraints) on minutes to talk. What are the constraints on the cell phone coverage for each company? Do they cover your area of the country well? Do they cover the entire United States well, or at least areas where you will be travelling? Is this coverage 5G, or is it less? Can you add a phone easily? Can you bring your previous phone number to this plan? The possibilities of constraints affecting each plan are several. So your task is to determine which plan is best, based on not only cost but also all constraints you deem important. Chapter Review Algebraic Expressions Linear Equations in One Variable with Applications Linear Inequalities in One Variable with Applications Ratio and Proportions Graphing Linear Equations and Inequalities Quadratic Equations In One Variable with Applications Functions Graphing Functions System of Linear Equations in Two Variables Systems of Linear Inequalities in Two Variables Linear Programming Chapter Test

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Introduction Financial health helps you realize your goals. (credit: modification of work "Budget and Bills" by Alabama Extension/Flickr, Public Domain) The topic of money management is a broad and sometimes complex one. Ultimately, personal money management involves managing both our debt and also our savings and investments. In 2021, the average American had consumer debt balance of $96,371. Nearly $100,000 per person. And less than 25% of Americans are debt free. Consumer debt can include mortgages, credit cards, as well as student loans. A key question all consumers should consider is how to manage debt and not become overburdened by it. The first step is to create a budget, which puts earnings into perspective, indicating what we can, and cannot, afford. A budget also entails setting aside certain funds for savings and investment, which help us achieve our short- and long-term goals. Creating a budget requires an understanding of how money—debt and savings—works. Initially, percentages and interest need to be understood. They drive most of what happens with debt and savings. With that understanding, discussions of buying a house, a car, or incurring credit card debt can be addressed from a financial perspective. All the while, retirement is waiting. Preparing for retirement involves saving and saving earlier rather than later. The power of compound interest is on full display when saving early. This chapter covers some of the basics of money management: percentages, interest, budgeting, debt (student loans, mortgage, car, credit cards), savings, investments, and taxes.

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Understanding Percent The federal budget describes how money is spent and how money is earned. (credit: "Breakdown of revenues and outlays in 2021 US Federal budget" Wikimedia Commons, Public Domain) Learning Objectives After completing this section, you should be able to: Define and calculate percent. Convert between percent, decimal, and fractional values. Calculate the total, percent, or part. Solve application problems involving percents. In 2020, the U.S. federal government budgeted $3.5 billion for the National Park Service , which appears to be a very large number (and is!) and a large portion of the total federal budget. However, the total outlays from the U.S. federal government in 2020 was $6.6 trillion . So, the amount budgeted for the National Park Service was less than one-tenth of 1 percent, or 1/10%, of the total outlays. This percent describes a specific number. Understanding that ratio puts the $3.5 billion budgeted to the National Park Service in perspective. This chapter focuses on percent as a primary tool for understanding money management. The interest paid on debt, the interest earned through investments, and even taxes are entirely determined using percent. This section introduces the basics of working with this invaluable tool. Define and Calculate Percent The word percent comes from the Latin phrase per centum, which means “by the hundred.” So any percent is a number divided by 100. Changing a percent to a fraction is to write the percent in its fractional form . To write n % in its fractional form is to write the percent as the fraction n 100 . A percent need not be an integer and does not have to be less than 100. Rewriting a Percent as a Fraction Rewrite the following as fractions: 18% 84% 38.7% 213% Using the definition and n = 18, 18% in fractional form is 18 100 . Using the definition and n = 84, 84% in fractional form is 84 100 . Using the definition and n = 38.7, 38.7% in fractional form is 38.7 100 . Using the definition and n = 213, 213% in fractional form is 213 100 . Convert Between Percent, Decimal, and Fractional Values When any calculation with a percent is to be performed, the form of the percent must be changed, either to its fractional form or its decimal form . We can change a percent into decimal form by dividing the percent by 100 and representing the result as a decimal. The decimal form of n % is found by calculating the decimal value of n ÷ 100 . Converting a Percent to Decimal Form Convert the following percents to decimal form: 17% 7% 18.45% To convert 17% to its decimal form, divide 17 by 100. This moves the decimal two places to the left, resulting in 0.17. The decimal form of 17% is 0.17. To convert 7% to its decimal form, divide 7 by 100. This moves the decimal two places to the left, resulting in 0.07. The decimal form of 7% is 0.07. To convert 18.45% to its decimal form, divide 18.45 by 100. This moves the decimal two places to the left, resulting in 0.1845. The decimal form of 18.45% is 0.1845. You should notice that, to convert from percent to decimal form, you can simply move the decimal two places to the left without performing the division. To convert the number x from decimal form to percent, multiply x by 100 and place a percent sign, %, after the number, ( x × 100 ) % . Converting the Decimal Form of a Percent to Percent Convert each of the following to percent: 0.34 4.15 0.0391 Using the formula and x = 0.34, we calculate ( 0.34 × 100 ) % , which gives us 34%. Using the formula and x = 4.15, we calculate ( 4.15 × 100 ) % , which gives us 415%. Using the formula and x = 0.0391, we calculate ( 0.0391 × 100 ) % , which gives us 3.91%. You should notice that, to convert from decimal form to percent form, you can simply move the decimal two places to the right without performing the multiplication. Calculate the Total, Percent, or Part The word “of” is used to indicate multiplication using fractions, as in “one-fourth of 56.” To find “one-fourth of 56” we would multiply 56 by one-fourth. We can think of percents as fractions with a specific denominator—100. So, to calculate “25% of 52,” we multiply 52 by 25%. But, first we need to convert the percent to either fractional form (25/100) or decimal form. Using the decimal form of 25% we have 0.25 × 52, which equals 13. In this problem, 52 is the total or base , 25 is the percentage , and 13 is the percentage of 52, or the part of 52. This is sometimes referred to as the amount . The mathematical formula relating the total (base), the percent in decimal form, and the part (amount) is part = percent × total , or, amount = percent × base . In all calculations, the percent is expressed in decimal form. Knowing any two of the values in our formula allows us to calculate the third value. In the following example, we know the total and the percent, and are asked to find the percentage of the total. Finding the Percent of a Total Determine 70% of 3,500 Determine 156% of 720 The total is x = 3,500, and the percent is n = 70. The decimal form of 70% is 0.70. To find the part, or percent of the total, substitute those values into the formula and calculate. part = percent × total = 0.70 × 3500 = 2450 From this, we say that 70% of 3,500 is 2,450. The total is x = 720, and the percent is n = 156. The decimal form of 156% is 1.56. To find the part, or percent of the total, substitute those values into the formula and calculate. part = percent × total = 1.56 × 720 = 1,123.2 From this, we say that 156% of 720 is 1,123.2. Finding Percent of a Total In the previous example, we knew the total and the percent and found the part using our formula. We may instead know the percent and the part, but not the total. We can use our formula again to solve for the total. Finding the Total from the Percent and the Part What is the total if 35% of the total is 70? What is the total if 10% of the total is 4,000? Step 1: The percent is 35, which in decimal form is 0.35. We were given that 35% of the total is 70, so the part is 70. We are to find the total. Substituting into the formula, we have part = percent × total 70 = 0.35 × total Step 2: To find the total, we solve the equation for the total. 70 = 0.35 × total 70 0.35 = 0.35 × total 0.35 200 = 0.35 × total 0.35 200 = total From this we see that 200 is the total, or, that 35% of 200 is 70. Step 1: The percent is 10, which in decimal form is 0.1. We were given that 10% of the total is 4,000, so the part is 4,000. Substituting into the formula, we have part = percent × total 4,000 = 0.1 × total Step 2: To find the total, we solve the equation for the total. 4,000 = 0.1 × total 4,000 0.1 = 0.1 × total 0.1 40,000 = 0.1 × total 0.1 40,000 = total From this we see that 40,000 is the total, or that 10% of 40,000 is 4,000. Finding the Total from the Percent and the Part Similarly, the percent can be found if the total and the percent of the total (the part) are known. This will result in the decimal form of the percent, so it must be converted to percent form. Finding the Percent from the Total and the Part What percent of 500 is 175? What percent of 228 is 155? Step 1: The total is 500, the percent of the total is 175. Substituting into the formula, we have part = percent × total 175 = percent × 500 Step 2: To find the percent, we solve the equation for the percent. 175 = percent × 500 175 500 = percent × 500 500 0.35 = percent × 500 500 0.35 = percent We see the percent in decimal form is 0.35. Converting from the decimal form yields 35%. We say that 175 is 35% of 500. Step 1: The total is 228, the percent of the total is 155. Substituting into the formula, we have part = percent × total 155 = percent × 228 Step 2: To find the percent, we solve the equation for the percent. 155 = percent × 228 155 228 = percent × 228 228 0.6798 = percent × 228 228 0.6798 = percent We see the percent is 0.6798 (rounded to four decimal places). Converting from the decimal form yields 67.98%. We say that 155 is 67.98% of 228. Finding the Percent When the Total and the Part Are Known Solve Application Problems Involving Percents Percents are frequently used in finance, research, science experiments, and even casual conversation. Understanding these types of values helps when consuming media or discussing finances, for instance. Effectively working with and interpreting numbers and percents will help you become an informed consumer of this information. In most cases, working through what is presented requires you to identify that you are indeed working with a question of percents, which two of the three values that are related through percents are known, and which of the three values you need to find. Retention Rate at College Justine applies to a medium size university outside her hometown and finds out that the retention rate (percent of students who return for their sophomore year) for the 2021 academic year at the university was 84%. During a visit to the registrar’s office, she finds out that 1,350 people had enrolled in academic year 2021. How many students from the academic year 2021 are returning for the 2022 academic year? The percent of students who will return for the 2022 academic year (the retention rate) is 84%. The total number of students who enrolled in the 2021 academic year was 1,350. This means the percent is known and the total is known. From this, we can determine the number of students who will return (percent of the total) for the 2022 academic year using the formula part = percent × total . Substituting into the formula and calculating, we find that the number of students that are returning is part = percent × total = 0.84 × 1,350 = 1,134 So 1,134 students will return for the 2022 academic year. Percent of Chemistry Majors Cameron enrolls in a calculus class. In this class of 45 students, there are 18 chemistry majors. What percent of the class are chemistry majors? In this situation, the percent is to be determined. We know the total number of students, 45, and the part of the students that are chemistry majors, 18. Using that information and the formula part = percent × total , the percent can be found. Substituting and solving, we have 18 = percent × 45 18 45 = percent × 45 45 0.4 = percent × 45 45 0.4 = percent Converting the 0.4 from decimal form, we find that 40% of the students in the calculus class are chemistry majors. Total Sales and Commission Mariel makes a 20% commission on every sale she makes. One week, her commission check is for $153.00. What were her total sales that week? In this problem, Mariel’s total sales is to be determined. We know the percent she earns is 20%. We also know that her sales commission was $153.00, which is the percent of the total. Using this information and the formula part = percent × total we can find Mariel’s total sales. The decimal form of 20% is 0.2. The part, or percent of the total, is 153. Substituting and solving, we obtain part = percent × total 153 = 0.2 × total 153 0.2 = 0.2 × total 0.2 765 = 0.2 × total 0.2 765 = total Mariel’s total sales were $765.00. LED Lightbulbs According to the energy website from the U.S. government , LED lightbulbs use at least 75% less energy than incandescent bulbs. They also last up to 25 times as long as an incandescent bulb. If lighting is a significant percent of your electrical use, replacing all incandescent bulbs with LED bulbs will significantly reduce your electric bill. Check Your Understanding For any answer, round to two decimal places, if necessary. Key Terms Percent Fractional form Decimal form Total Base Percent of the total Part Amount Key Concepts Know what a percent is as a fraction, a decimal, and as a part of the whole. Use the percent equation to find any of the three values that are related by the equation. Apply the percent equation in applications. Video Finding Percent of a Total Finding the Total from the Percent and the Part Finding the Percent When the Total and the Part Are Known Formulas part = percent x total

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