---
source: ncert_exercises
topic: algebra
class_level: class_10
chapter: ch3
exercise: ex3.3
difficulty: intermediate
---

Class 10 | Ch3: Pair of Linear Equations | Exercise 3.3 (Substitution Method)

Q1. Solve by substitution:
(i) x+y=14, x-y=4  (ii) s-t=3, s/3+t/2=6  (iii) 3x-y=3, 9x-3y=9
(iv) 0.2x+0.3y=1.3, 0.4x+0.5y=2.3  (v) √2x+√3y=0, √3x-√8y=0
(vi) 3x/2-5y/3=-2, x/3+y/2=13/6
Answer:
(i) From x-y=4: x=y+4. Sub in x+y=14: 2y+4=14 → y=5, x=9. ✅
(ii) s=t+3. Sub: (t+3)/3+t/2=6 → 2t+6+3t=36 → t=6, s=9. ✅
(iii) From 3x-y=3: y=3x-3. Sub in 9x-3y=9: 9x-9x+9=9 → 9=9. Infinite solutions. ✅
(iv) x=2.5, y=2.5. ✅
(v) From √2x=-√3y: x=-√3y/√2. Sub: √3(-√3y/√2)-√8y=0 → -3y/√2-2√2y=0 → y(-3/√2-2√2)=0 → y=0, x=0. ✅
(vi) x=2, y=3. ✅

Q2. Solve: 2x+3y=11 and 2x-4y=-24. Find m if y=mx+3.
Answer: Subtract: 7y=35 → y=5. 2x=11-15=-4 → x=-2. y=mx+3: 5=-2m+3 → m=-1. ✅

Q3. Form equations and solve by substitution:
(i) Two numbers differ by 26, one is 3 times other.
(ii) Larger of supplementary angles exceeds smaller by 18°.
(iii) 7 bats+6 balls=₹3800, 3 bats+5 balls=₹1750.
(iv) Fixed taxi charge + per km charge. 10km=₹105, 15km=₹155.
Answer:
(i) x=3y, x-y=26. 2y=26 → y=13, x=39. ✅
(ii) x+y=180, x-y=18. x=99°, y=81°. ✅
(iii) 7x+6y=3800, 3x+5y=1750. x=500 (bat), y=50 (ball). ✅
(iv) x+10y=105, x+15y=155. y=10 (per km), x=5 (fixed). For 25 km: 5+25×10=₹255. ✅