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"original_text": "3x+7=22",
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"problem_text": "3x + 7 = 22",
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"solution": "1. **Understanding the problem:**\nThe problem is to find the value of the variable 'x' that makes the equation \"3x + 7 = 22\" true. This is a linear equation with one variable, and we need to isolate 'x' to find its value.\n\n2. **Step-by-step solution:**\nWe will use algebraic manipulation to solve for 'x'. The goal is to get 'x' by itself on one side of the equation.\n\n* **Step 1: Subtract 7 from both sides of the equation.**\n The equation is: 3x + 7 = 22\n To isolate the term with 'x' (which is 3x), we need to remove the '+7'. We do this by performing the inverse operation, which is subtraction. We must subtract 7 from both sides of the equation to maintain equality.\n 3x + 7 - 7 = 22 - 7\n This simplifies to:\n 3x = 15\n\n* **Step 2: Divide both sides of the equation by 3.**\n Now we have: 3x = 15\n To isolate 'x', we need to undo the multiplication by 3. We do this by performing the inverse operation, which is division. We must divide both sides of the equation by 3.\n 3x / 3 = 15 / 3\n This simplifies to:\n x = 5\n\n3. **Final Answer:**\nThe value of 'x' that satisfies the equation 3x + 7 = 22 is 5.\n\nWe can check our answer by substituting x = 5 back into the original equation:\n3(5) + 7 = 15 + 7 = 22.\nSince 22 = 22, our solution is correct.",
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"original_text": "a=(2,\u22121,3),b=(1,4,2)",
"parsed_question": {
"problem_text": "a = (2, -1, 3), b = (1, 4, 2)",
"topic": "linear_algebra",
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"solution": "### 1. Understanding of the Problem\n\nThe problem provides two vectors, $\\mathbf{a} = (2, -1, 3)$ and $\\mathbf{b} = (1, 4, 2)$. The topic is linear algebra and the strategy is matrix operations. While the problem doesn't explicitly state what operation to perform, in the context of linear algebra with two vectors, common operations include vector addition, subtraction, scalar multiplication, dot product, and cross product. Given the instruction to use \"matrix operations\", it implies we might need to represent these vectors as matrices or perform operations that can be viewed through a matrix lens. Without a specific question, I will assume the most common vector operation asked in conjunction with linear algebra and matrix operations is to find the dot product of the two vectors. The dot product of two vectors can be computed using matrix multiplication if the vectors are represented as row or column matrices.\n\n### 2. Step-by-Step Solution\n\nTo calculate the dot product of vectors $\\mathbf{a}$ and $\\mathbf{b}$ using matrix operations, we can represent them as column matrices and perform matrix multiplication.\n\n**Step 1: Represent the vectors as column matrices.**\nLet $\\mathbf{a}$ be represented by a column matrix $A$ and $\\mathbf{b}$ be represented by a column matrix $B$.\n$A = \\begin{pmatrix} 2 \\\\ -1 \\\\ 3 \\end{pmatrix}$\n$B = \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix}$\n\n**Step 2: Calculate the dot product using matrix multiplication.**\nThe dot product of $\\mathbf{a}$ and $\\mathbf{b}$ can be calculated as $A^T B$, where $A^T$ is the transpose of matrix $A$. The transpose of a column matrix is a row matrix.\n$A^T = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix}$\n\nNow, perform the matrix multiplication:\n$A^T B = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix}$\n\nTo multiply these matrices, we multiply each element of the first matrix (the row vector) by the corresponding element of the second matrix (the column vector) and then sum the results.\n\n$(2 \\times 1) + (-1 \\times 4) + (3 \\times 2)$\n\n**Step 3: Perform the arithmetic operations.**\n$2 \\times 1 = 2$\n$-1 \\times 4 = -4$\n$3 \\times 2 = 6$\n\nNow, sum these values:\n$2 + (-4) + 6 = 2 - 4 + 6 = -2 + 6 = 4$\n\nAlternatively, we could represent $\\mathbf{a}$ as a row matrix and $\\mathbf{b}$ as a column matrix, or $\\mathbf{a}$ as a column matrix and $\\mathbf{b}$ as a row matrix for a different matrix operation that results in the dot product. The standard way to obtain a scalar (the dot product) from two vectors using matrix multiplication is to multiply a row vector by a column vector.\n\nLet's consider representing $\\mathbf{a}$ as a row vector and $\\mathbf{b}$ as a column vector.\n$\\mathbf{a}^T = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix}$\n$\\mathbf{b} = \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix}$\nDot product = $\\mathbf{a}^T \\mathbf{b} = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix} = (2 \\times 1) + (-1 \\times 4) + (3 \\times 2) = 2 - 4 + 6 = 4$.\n\nOr, representing $\\mathbf{a}$ as a column vector and $\\mathbf{b}$ as a row vector (transpose of $\\mathbf{b}$), and then multiplying $\\mathbf{a}$ by $\\mathbf{b}^T$:\n$\\mathbf{a} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 3 \\end{pmatrix}$\n$\\mathbf{b}^T = \\begin{pmatrix} 1 & 4 & 2 \\end{pmatrix}$\nDot product = $\\mathbf{a} \\mathbf{b}^T = \\begin{pmatrix} 2 \\\\ -1 \\\\ 3 \\end{pmatrix} \\begin{pmatrix} 1 & 4 & 2 \\end{pmatrix}$. This would result in a $3 \\times 3$ matrix, not a scalar dot product. The dot product is a scalar quantity.\n\nThe correct matrix operation to obtain the dot product is the multiplication of a row vector with a column vector.\n\nLet $\\mathbf{a} = (a_1, a_2, a_3)$ and $\\mathbf{b} = (b_1, b_2, b_3)$.\nThe dot product is $\\mathbf{a} \\cdot \\mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.\n\nUsing matrix notation:\nRepresent $\\mathbf{a}$ as a row matrix $A_{row} = \\begin{pmatrix} a_1 & a_2 & a_3 \\end{pmatrix}$ and $\\mathbf{b}$ as a column matrix $B_{col} = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix}$.\nThen $\\mathbf{a} \\cdot \\mathbf{b} = A_{row} B_{col} = \\begin{pmatrix} a_1 & a_2 & a_3 \\end{pmatrix} \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix} = a_1 b_1 + a_2 b_2 + a_3 b_3$.\n\nSubstituting the given values:\n$A_{row} = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix}$\n$B_{col} = \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix}$\n\nDot product = $\\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix} = (2)(1) + (-1)(4) + (3)(2) = 2 - 4 + 6 = 4$.\n\n### 3. Final Answer\n\nThe dot product of vectors $\\mathbf{a}$ and $\\mathbf{b}$ can be computed using matrix operations as follows:\nLet $\\mathbf{a}$ be represented by the row matrix $A_{row} = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix}$ and $\\mathbf{b}$ be represented by the column matrix $B_{col} = \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix}$.\n\nThe dot product is then calculated by the matrix multiplication $A_{row} B_{col}$:\n$A_{row} B_{col} = \\begin{pmatrix} 2 & -1 & 3 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 4 \\\\ 2 \\end{pmatrix} = (2 \\times 1) + (-1 \\times 4) + (3 \\times 2) = 2 - 4 + 6 = 4$.\n\nThe final answer is $\\boxed{4}$.",
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"feedback": "The solution correctly identifies the likely intended operation (dot product) given the problem statement and context. It clearly explains the matrix representation of the vectors and demonstrates the matrix multiplication process to compute the dot product. The arithmetic is also correct."
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"timestamp": "2026-01-17T20:41:17.697928",
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"original_text": "x**2 +5x+1=0",
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"problem_text": "x^2 + 5x + 1 = 0",
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"solution": "1. **Understanding the problem:**\nThe problem asks us to solve the quadratic equation $x^2 + 5x + 1 = 0$ for the variable $x$. This is an equation of the form $ax^2 + bx + c = 0$.\n\n2. **Step-by-step solution:**\nWe will use the quadratic formula to solve this equation, as it is a standard method for finding the roots of quadratic equations. The quadratic formula is given by:\n$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$\n\nIn our equation, $x^2 + 5x + 1 = 0$, we have:\n$a = 1$\n$b = 5$\n$c = 1$\n\nFirst, let's calculate the discriminant, $D = b^2 - 4ac$:\nCALCULATE: $5^2 - 4 * 1 * 1$\n \u2192 Calculation error: Cannot evaluate: $5**2 - 4 * 1 * 1$\n$D = 25 - 4$\n$D = 21$\n\nSince the discriminant $D = 21$ is greater than 0, there will be two distinct real roots.\n\nNow, we can substitute the values of $a$, $b$, and $D$ into the quadratic formula:\n$x = \\frac{-5 \\pm \\sqrt{21}}{2 * 1}$\n$x = \\frac{-5 \\pm \\sqrt{21}}{2}$\n\nThis gives us two possible solutions for $x$:\n\nSolution 1:\n$x_1 = \\frac{-5 + \\sqrt{21}}{2}$\n\nSolution 2:\n$x_2 = \\frac{-5 - \\sqrt{21}}{2}$\n\n3. **Final Answer:**\nThe solutions to the equation $x^2 + 5x + 1 = 0$ are $x = \\frac{-5 + \\sqrt{21}}{2}$ and $x = \\frac{-5 - \\sqrt{21}}{2}$.",
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"content": "ALGEBRA FORMULAS AND CONCEPTS\n\nQuadratic Equations:\n- Standard form: ax\u00b2 + bx + c = 0\n- Quadratic formula: x = (-b \u00b1 \u221a(b\u00b2-4ac)) / 2a\n- Discriminant: D = b\u00b2 - 4ac\n - D > 0: two real roots\n - D = 0: one real root\n - D < 0: complex roots\n- Sum of roots: -b/a\n- Product of roots: c/a\n\nPolynomial Identities:\n- (a+b)\u00b2 = a\u00b2 + 2ab + b\u00b2\n- (a-b)\u00b2 = a\u00b2 - 2ab + b\u00b2\n- a\u00b2 - b\u00b2 = (a+b)(a-b)\n- (a+b)\u00b3 = a\u00b3 + 3a\u00b2b + 3ab\u00b2 + b\u00b3\n- (a-b)\u00b3 = a\u00b3 - 3a\u00b2b + 3ab\u00b2 - b\u00b3\n- a\u00b3 + b\u00b3 = (a+b)(a\u00b2 - ab + b\u00b2)\n- a\u00b3 - b\u00b3 = (a-b)(a\u00b2 + ab + b\u00b2)\n",
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"content": "Inequalities:\n- |x| < a \u27f9 -a < x < a\n- |x| > a \u27f9 x < -a or x > a\n- AM \u2265 GM: (a+b)/2 \u2265 \u221a(ab) for a,b > 0\n\nCommon Mistakes:\n- Don't forget to check domain restrictions\n- Remember to verify solutions in original equation\n- Check for extraneous solutions after squaring",
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"solution": "1. **Understanding the problem:**\nThe problem is to find the value of the variable 'x' that makes the equation \"3x + 7 = 22\" true. This is a linear equation with one variable, and we need to isolate 'x' to find its value.\n\n2. **Step-by-step solution:**\nWe will use algebraic manipulation to solve for 'x'. The goal is to get 'x' by itself on one side of the equation.\n\n* **Step 1: Subtract 7 from both sides of the equation.**\n The equation is: 3x + 7 = 22\n To isolate the term with 'x' (which is 3x), we need to remove the '+7'. We do this by performing the inverse operation, which is subtraction. We must subtract 7 from both sides of the equation to maintain equality.\n 3x + 7 - 7 = 22 - 7\n This simplifies to:\n 3x = 15\n\n* **Step 2: Divide both sides of the equation by 3.**\n Now we have: 3x = 15\n To isolate 'x', we need to undo the multiplication by 3. We do this by performing the inverse operation, which is division. We must divide both sides of the equation by 3.\n 3x / 3 = 15 / 3\n This simplifies to:\n x = 5\n\n3. **Final Answer:**\nThe value of 'x' that satisfies the equation 3x + 7 = 22 is 5.\n\nWe can check our answer by substituting x = 5 back into the original equation:\n3(5) + 7 = 15 + 7 = 22.\nSince 22 = 22, our solution is correct.",
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"original_text": "Find the derivatives of x squared plus 3x.",
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"problem_text": "Find the derivatives of x squared plus 3x.",
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"solution": "**1. Understanding of the Problem**\n\nThe problem asks us to find the derivative of the function $f(x) = x^2 + 3x$. This involves applying the basic rules of differentiation.\n\n**2. Step-by-Step Solution**\n\n* **Step 1: Identify the function.**\n The function we need to differentiate is $f(x) = x^2 + 3x$.\n\n* **Step 2: Apply the sum rule for differentiation.**\n The sum rule states that the derivative of a sum of functions is the sum of their derivatives: $d/dx[f(x) + g(x)] = d/dx[f(x)] + d/dx[g(x)]$.\n In our case, $f(x) = x^2$ and $g(x) = 3x$. So, $d/dx(x^2 + 3x) = d/dx(x^2) + d/dx(3x)$.\n\n* **Step 3: Differentiate the first term, $x^2$.**\n We will use the power rule, which states $d/dx(x^n) = nx^(n-1)$.\n For $x^2$, $n=2$.\n CALCULATE: 2 * x**(2-1)\n \u2192 Calculation error: Cannot evaluate: 2 * x**(2-1)\n This gives us $2x^1$, which simplifies to $2x$.\n\n* **Step 4: Differentiate the second term, $3x$.**\n We can consider this as a constant multiple of $x$. The constant multiple rule states $d/dx(c \\cdot f(x)) = c \\cdot d/dx(f(x))$. Here, $c=3$ and $f(x)=x$.\n The derivative of $x$ is 1 (using the power rule with $n=1$: $1 \\cdot x^(1-1) = 1 \\cdot x^0 = 1$).\n CALCULATE: 3 * 1\n \u2192 Result: 3\n This gives us $3$.\n\n* **Step 5: Combine the derivatives of the terms.**\n Now we add the derivatives from Step 3 and Step 4.\n CALCULATE: 2*x + 3\n \u2192 Calculation error: Cannot evaluate: 2*x + 3\n So, the derivative of $x^2 + 3x$ is $2x + 3$.\n\n**3. Final Answer**\n\nThe derivative of $x^2 + 3x$ is $2x + 3$.",
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"content": "- d/dx(sin x) = cos x\n- d/dx(cos x) = -sin x\n- d/dx(tan x) = sec\u00b2x\n\nOptimization:\n- Critical points: f'(x) = 0\n- Local max: f'(x) = 0 and f''(x) < 0\n- Local min: f'(x) = 0 and f''(x) > 0\n\nCommon Mistakes:\n- Forgetting chain rule\n- Sign errors in derivatives\n- Not checking endpoints in optimization",
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"content": "CALCULUS FORMULAS AND CONCEPTS\n\nLimits:\n- lim(x\u2192a) [f(x) + g(x)] = lim(x\u2192a) f(x) + lim(x\u2192a) g(x)\n- lim(x\u2192a) [f(x) \u00d7 g(x)] = lim(x\u2192a) f(x) \u00d7 lim(x\u2192a) g(x)\n- lim(x\u21920) (sin x)/x = 1\n- lim(x\u2192\u221e) (1 + 1/x)^x = e\n\nL'H\u00f4pital's Rule:\n- For 0/0 or \u221e/\u221e: lim f(x)/g(x) = lim f'(x)/g'(x)\n\nDerivatives:\n- Power rule: d/dx(x^n) = nx^(n-1)\n- Product rule: d/dx(uv) = u'v + uv'\n- Quotient rule: d/dx(u/v) = (u'v - uv')/v\u00b2\n- Chain rule: d/dx[f(g(x))] = f'(g(x)) \u00d7 g'(x)\n\nStandard Derivatives:\n- d/dx(e^x) = e^x\n- d/dx(ln x) = 1/x",
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"original_text": "x**2 +5x+1=0",
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"problem_text": "x^2 + 5x + 1 = 0",
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"solution": "1. **Understanding the problem:**\nThe problem asks us to find the values of $x$ that satisfy the quadratic equation $x^2 + 5x + 1 = 0$. This equation is in the standard quadratic form $ax^2 + bx + c = 0$, where $a=1$, $b=5$, and $c=1$.\n\n2. **Step-by-step solution:**\nWe will use the quadratic formula to solve for $x$. The quadratic formula is given by $x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$.\n\n**Step 1: Identify the coefficients a, b, and c.**\nFrom the equation $x^2 + 5x + 1 = 0$, we have:\n$a = 1$\n$b = 5$\n$c = 1$\n\n**Step 2: Calculate the discriminant ($\\Delta = b^2 - 4ac$).**\nThe discriminant will tell us the nature of the roots.\nCALCULATE: $5^2 - 4 * 1 * 1$\n \u2192 Calculation error: Cannot evaluate: $5**2 - 4 * 1 * 1$\nCALCULATE: $25 - 4$\n \u2192 Calculation error: Cannot evaluate: $25 - 4$\nCALCULATE: $21$\n \u2192 Calculation error: Cannot evaluate: $21$\nSince the discriminant is $21$, which is greater than 0, there will be two distinct real roots.\n\n**Step 3: Apply the quadratic formula.**\nNow, substitute the values of $a$, $b$, and the discriminant into the quadratic formula:\n$x = \\frac{-b \\pm \\sqrt{\\Delta}}{2a}$\n$x = \\frac{-5 \\pm \\sqrt{21}}{2 * 1}$\n$x = \\frac{-5 \\pm \\sqrt{21}}{2}$\n\n**Step 4: Write out the two distinct solutions.**\nThe two solutions are:\n$x_1 = \\frac{-5 + \\sqrt{21}}{2}$\n$x_2 = \\frac{-5 - \\sqrt{21}}{2}$\n\n3. **Final Answer:**\nThe solutions to the equation $x^2 + 5x + 1 = 0$ are $x = \\frac{-5 + \\sqrt{21}}{2}$ and $x = \\frac{-5 - \\sqrt{21}}{2}$.",
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"content": "ALGEBRA FORMULAS AND CONCEPTS\n\nQuadratic Equations:\n- Standard form: ax\u00b2 + bx + c = 0\n- Quadratic formula: x = (-b \u00b1 \u221a(b\u00b2-4ac)) / 2a\n- Discriminant: D = b\u00b2 - 4ac\n - D > 0: two real roots\n - D = 0: one real root\n - D < 0: complex roots\n- Sum of roots: -b/a\n- Product of roots: c/a\n\nPolynomial Identities:\n- (a+b)\u00b2 = a\u00b2 + 2ab + b\u00b2\n- (a-b)\u00b2 = a\u00b2 - 2ab + b\u00b2\n- a\u00b2 - b\u00b2 = (a+b)(a-b)\n- (a+b)\u00b3 = a\u00b3 + 3a\u00b2b + 3ab\u00b2 + b\u00b3\n- (a-b)\u00b3 = a\u00b3 - 3a\u00b2b + 3ab\u00b2 - b\u00b3\n- a\u00b3 + b\u00b3 = (a+b)(a\u00b2 - ab + b\u00b2)\n- a\u00b3 - b\u00b3 = (a-b)(a\u00b2 + ab + b\u00b2)\n",
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"solution": "1. **Understanding the problem:**\nThe problem asks us to solve the quadratic equation $x^2 + 5x + 1 = 0$ for the variable $x$. This is an equation of the form $ax^2 + bx + c = 0$.\n\n2. **Step-by-step solution:**\nWe will use the quadratic formula to solve this equation, as it is a standard method for finding the roots of quadratic equations. The quadratic formula is given by:\n$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$\n\nIn our equation, $x^2 + 5x + 1 = 0$, we have:\n$a = 1$\n$b = 5$\n$c = 1$\n\nFirst, let's calculate the discriminant, $D = b^2 - 4ac$:\nCALCULATE: $5^2 - 4 * 1 * 1$\n \u2192 Calculation error: Cannot evaluate: $5**2 - 4 * 1 * 1$\n$D = 25 - 4$\n$D = 21$\n\nSince the discriminant $D = 21$ is greater than 0, there will be two distinct real roots.\n\nNow, we can substitute the values of $a$, $b$, and $D$ into the quadratic formula:\n$x = \\frac{-5 \\pm \\sqrt{21}}{2 * 1}$\n$x = \\frac{-5 \\pm \\sqrt{21}}{2}$\n\nThis gives us two possible solutions for $x$:\n\nSolution 1:\n$x_1 = \\frac{-5 + \\sqrt{21}}{2}$\n\nSolution 2:\n$x_2 = \\frac{-5 - \\sqrt{21}}{2}$\n\n3. **Final Answer:**\nThe solutions to the equation $x^2 + 5x + 1 = 0$ are $x = \\frac{-5 + \\sqrt{21}}{2}$ and $x = \\frac{-5 - \\sqrt{21}}{2}$.",
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"content": "ALGEBRA FORMULAS AND CONCEPTS\n\nQuadratic Equations:\n- Standard form: ax\u00b2 + bx + c = 0\n- Quadratic formula: x = (-b \u00b1 \u221a(b\u00b2-4ac)) / 2a\n- Discriminant: D = b\u00b2 - 4ac\n - D > 0: two real roots\n - D = 0: one real root\n - D < 0: complex roots\n- Sum of roots: -b/a\n- Product of roots: c/a\n\nPolynomial Identities:\n- (a+b)\u00b2 = a\u00b2 + 2ab + b\u00b2\n- (a-b)\u00b2 = a\u00b2 - 2ab + b\u00b2\n- a\u00b2 - b\u00b2 = (a+b)(a-b)\n- (a+b)\u00b3 = a\u00b3 + 3a\u00b2b + 3ab\u00b2 + b\u00b3\n- (a-b)\u00b3 = a\u00b3 - 3a\u00b2b + 3ab\u00b2 - b\u00b3\n- a\u00b3 + b\u00b3 = (a+b)(a\u00b2 - ab + b\u00b2)\n- a\u00b3 - b\u00b3 = (a-b)(a\u00b2 + ab + b\u00b2)\n",
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"content": "Inequalities:\n- |x| < a \u27f9 -a < x < a\n- |x| > a \u27f9 x < -a or x > a\n- AM \u2265 GM: (a+b)/2 \u2265 \u221a(ab) for a,b > 0\n\nCommon Mistakes:\n- Don't forget to check domain restrictions\n- Remember to verify solutions in original equation\n- Check for extraneous solutions after squaring",
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"solution": "1. **Understanding the problem:**\nThe problem is to find the value of the variable 'x' that makes the equation \"3x + 7 = 22\" true. This is a linear equation with one variable, and we need to isolate 'x' to find its value.\n\n2. **Step-by-step solution:**\nWe will use algebraic manipulation to solve for 'x'. The goal is to get 'x' by itself on one side of the equation.\n\n* **Step 1: Subtract 7 from both sides of the equation.**\n The equation is: 3x + 7 = 22\n To isolate the term with 'x' (which is 3x), we need to remove the '+7'. We do this by performing the inverse operation, which is subtraction. We must subtract 7 from both sides of the equation to maintain equality.\n 3x + 7 - 7 = 22 - 7\n This simplifies to:\n 3x = 15\n\n* **Step 2: Divide both sides of the equation by 3.**\n Now we have: 3x = 15\n To isolate 'x', we need to undo the multiplication by 3. We do this by performing the inverse operation, which is division. We must divide both sides of the equation by 3.\n 3x / 3 = 15 / 3\n This simplifies to:\n x = 5\n\n3. **Final Answer:**\nThe value of 'x' that satisfies the equation 3x + 7 = 22 is 5.\n\nWe can check our answer by substituting x = 5 back into the original equation:\n3(5) + 7 = 15 + 7 = 22.\nSince 22 = 22, our solution is correct.",
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"solution": "1. **Understanding the problem:**\nThe problem is to find the value of the variable 'x' that makes the equation \"3x + 7 = 22\" true. This is a linear equation with one variable, and we need to isolate 'x' to find its value.\n\n2. **Step-by-step solution:**\nWe will use algebraic manipulation to solve for 'x'. The goal is to get 'x' by itself on one side of the equation.\n\n* **Step 1: Subtract 7 from both sides of the equation.**\n The equation is: 3x + 7 = 22\n To isolate the term with 'x' (which is 3x), we need to remove the '+7'. We do this by performing the inverse operation, which is subtraction. We must subtract 7 from both sides of the equation to maintain equality.\n 3x + 7 - 7 = 22 - 7\n This simplifies to:\n 3x = 15\n\n* **Step 2: Divide both sides of the equation by 3.**\n Now we have: 3x = 15\n To isolate 'x', we need to undo the multiplication by 3. We do this by performing the inverse operation, which is division. We must divide both sides of the equation by 3.\n 3x / 3 = 15 / 3\n This simplifies to:\n x = 5\n\n3. **Final Answer:**\nThe value of 'x' that satisfies the equation 3x + 7 = 22 is 5.\n\nWe can check our answer by substituting x = 5 back into the original equation:\n3(5) + 7 = 15 + 7 = 22.\nSince 22 = 22, our solution is correct.",
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