Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.4 NCERT Solutions

Introduction

Exercise 3.4 focuses on solving pairs of linear equations using the elimination method. In this method, coefficients of one variable are made equal, then equations are added or subtracted to eliminate that variable. This exercise builds the foundation for solving real‑life problems algebraically.

Formula Used

• Elimination Method Steps:

  1. Multiply equations (if needed) to make coefficients of one variable equal.

  2. Add or subtract equations to eliminate one variable.

  3. Solve for the remaining variable.

  4. Substitute back to find the other variable.

NCERT Questions with Solutions (10)

Q1. Solve: $2x+3y=11$, $4x+6y=22$.

Solution: Both equations are multiples ⇒ infinitely many solutions.

Q2. Solve: $3x+2y=16$, $2x-3y=-6$.

Solution: Multiply 1st by 3, 2nd by 2: $9x+6y=48$, $4x-6y=-12$. Add: $13x=36\Rightarrow x=\frac{36}{13}$. Substitute: $y=\frac{18}{13}$. Solution: $(\frac{36}{13},\frac{18}{13})$.

Q3. Solve: $x+y=5$, $x-y=1$.

Solution: Add: $2x=6\Rightarrow x=3$. Then $y=2$. Solution: $(3,2)$.

Q4. Solve: $2x+y=10$, $x-y=2$.

Solution: Multiply 2nd by 2: $2x-2y=4$. Subtract from 1st: $3y=6\Rightarrow y=2$. Then $x=4$. Solution: $(4,2)$.

Q5. Solve: $3x+4y=20$, $5x-2y=10$.

Solution: Multiply 1st by 2: $6x+8y=40$. Multiply 2nd by 4: $20x-8y=40$. Add: $26x=80\Rightarrow x=\frac{40}{13}$. Then $y=\frac{35}{13}$. Solution: $(\frac{40}{13},\frac{35}{13})$.

Q6. Solve: $x+2y=7$, $3x-y=5$.

Solution: Multiply 1st by 3: $3x+6y=21$. Subtract 2nd: $7y=16\Rightarrow y=\frac{16}{7}$. Then $x=\frac{15}{7}$. Solution: $(\frac{15}{7},\frac{16}{7})$.

Q7. Solve: $2x+3y=12$, $x-y=1$.

Solution: Multiply 2nd by 3: $3x-3y=3$. Add: $5x=15\Rightarrow x=3$. Then $y=2$. Solution: $(3,2)$.

Q8. Solve: $x+y=9$, $x-2y=4$.

Solution: Subtract: $3y=5\Rightarrow y=\frac{5}{3}$. Then $x=\frac{22}{3}$. Solution: $(\frac{22}{3},\frac{5}{3})$.

Q9. Solve: $3x+y=10$, $2x-y=2$.

Solution: Add: $5x=12\Rightarrow x=\frac{12}{5}$. Then $y=\frac{14}{5}$. Solution: $(\frac{12}{5},\frac{14}{5})$.

Q10. Solve: $x+3y=6$, $2x-y=3$.

Solution: Multiply 2nd by 3: $6x-3y=9$. Add with 1st: $7x=15\Rightarrow x=\frac{15}{7}$. Then $y=\frac{9}{7}$. Solution: $(\frac{15}{7},\frac{9}{7})$.

FAQs (10 from NCERT)

1. Q: What is the elimination method? A: A method of solving equations by eliminating one variable.

2. Q: What is the first step in elimination method? A: Make coefficients of one variable equal.

3. Q: What is the second step in elimination method? A: Add or subtract equations to eliminate that variable.

4. Q: What is the third step in elimination method? A: Solve for the remaining variable.

5. Q: What is the fourth step in elimination method? A: Substitute back to find the other variable.

6. Q: What is a consistent pair of equations? A: A pair that has at least one solution.

7. Q: What is an inconsistent pair of equations? A: A pair that has no solution.

8. Q: What is a dependent pair of equations? A: A pair that has infinitely many solutions.

9. Q: Why is elimination method useful? A: It is systematic and works well for integer coefficients.

10. Q: Why is this exercise important? A: It builds algebraic skills for solving real‑life problems.

Conclusion

Exercise 3.4 has 10 solved questions and 10 FAQs that strengthen your understanding of solving pairs of linear equations using the elimination method. This builds the foundation for algebraic problem‑solving in Class 10 Maths.

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