---
source: ncert_exercises
topic: algebra
class_level: class_9
chapter: ch4
exercise: ex4.3
difficulty: intermediate
---

Class 9 | Ch4: Linear Equations in Two Variables | Exercise 4.3

Key: Graph of a linear equation ax+by+c=0 is a straight line.
To draw: find at least 2 points satisfying the equation, plot and join.

Q1. Draw graph of: (i) x+y=4  (ii) x-y=2  (iii) y=3x  (iv) 3=2x+y
Answer:
(i) x+y=4: Points: (0,4), (4,0), (2,2). Draw line through these. ✅
(ii) x-y=2: Points: (0,-2), (2,0), (4,2). Draw line through these. ✅
(iii) y=3x: Points: (0,0), (1,3), (-1,-3). Passes through origin. ✅
(iv) 2x+y=3: Points: (0,3), (3/2,0), (1,1). Draw line. ✅

Q2. Give equations of two lines passing through (2,14). How many such lines can there be?
Answer: Two possible equations: x+y=16 (since 2+14=16) and 7x-y=0 (since 14=7×2).
Infinitely many lines pass through any given point. ✅

Q3. If the point (3,4) lies on graph of 3y=ax+7, find value of a.
Answer: (3,4) satisfies equation: 3(4)=a(3)+7 → 12=3a+7 → 3a=5 → a=5/3. ✅

Q4. Taxi fare in a city: for first km ₹8, for subsequent km ₹5/km. Taking distance=x km and fare=₹y, write linear equation. Draw graph.
Answer: y = 8 + 5(x-1) = 8 + 5x - 5 = 5x + 3. Equation: y = 5x + 3 or 5x - y + 3 = 0.
Points for graph: (0,3), (1,8), (2,13). ✅

Q5. Yamini and Fatima together contributed ₹100 towards the Prime Minister's Relief Fund. Write linear equation and draw graph.
Answer: Let Yamini contribute ₹x, Fatima ₹y. x+y=100.
Points: (0,100), (100,0), (50,50). ✅

Q6. In countries like USA, temperature is in Fahrenheit (°F). India uses Celsius (°C). F=(9/5)C+32. Draw graph. From graph: (i) 30°C=? °F (ii) 95°F=?°C (iii) temperature same in both scales.
Answer: Equation: F=(9/5)C+32. Points: C=0→F=32: (0,32). C=5→F=41: (5,41). C=10→F=50: (10,50).
(i) C=30: F=9/5(30)+32=54+32=86°F. ✅
(ii) F=95: 95=9C/5+32 → 63=9C/5 → C=35°C. ✅
(iii) F=C: C=9C/5+32 → -4C/5=32 → C=-40. So at -40° both scales are equal. ✅