Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.3 NCERT Solutions

Introduction

Exercise 3.3 focuses on solving pairs of linear equations using the substitution method. In this method, one equation is solved for one variable, and the value is substituted into the other equation to find the solution. This exercise builds the foundation for algebraic methods of solving equations.

Formula Used

• Substitution Method Steps:

  1. Solve one equation for one variable.

  2. Substitute this value into the other equation.

  3. Solve for the second variable.

  4. Substitute back to find the first variable.

NCERT Questions with Solutions (10)

Q1. Solve: $x+y=14$, $x-y=4$.

Solution: From 2nd equation: $x=y+4$. Substitute in 1st: $y+4+y=14\Rightarrow 2y=10\Rightarrow y=5$. Then $x=9$. Solution: $(9,5)$.

Q2. Solve: $2x+3y=11$, $2x-4y=-6$.

Solution: From 2nd equation: $2x=4y-6\Rightarrow x=2y-3$. Substitute in 1st: $2(2y-3)+3y=11\Rightarrow 4y-6+3y=11\Rightarrow 7y=17\Rightarrow y=\frac{17}{7}$. Then $x=2(\frac{17}{7})-3=\frac{34}{7}-3=\frac{13}{7}$. Solution: $(\frac{13}{7},\frac{17}{7})$.

Q3. Solve: $x+2y=7$, $3x-y=5$.

Solution: From 1st: $x=7-2y$. Substitute in 2nd: $3(7-2y)-y=5\Rightarrow 21-6y-y=5\Rightarrow -7y=-16\Rightarrow y=\frac{16}{7}$. Then $x=7-2(\frac{16}{7})=\frac{15}{7}$. Solution: $(\frac{15}{7},\frac{16}{7})$.

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

Solution: From 1st: $x=y+2$. Substitute in 2nd: $y+2+y=10\Rightarrow 2y=8\Rightarrow y=4$. Then $x=6$. Solution: $(6,4)$.

Q5. Solve: $2x+y=11$, $x-y=3$.

Solution: From 2nd: $x=y+3$. Substitute in 1st: $2(y+3)+y=11\Rightarrow 2y+6+y=11\Rightarrow 3y=5\Rightarrow y=\frac{5}{3}$. Then $x=\frac{5}{3}+3=\frac{14}{3}$. Solution: $(\frac{14}{3},\frac{5}{3})$.

Q6. Solve: $3x+2y=12$, $x-y=1$.

Solution: From 2nd: $x=y+1$. Substitute in 1st: $3(y+1)+2y=12\Rightarrow 3y+3+2y=12\Rightarrow 5y=9\Rightarrow y=\frac{9}{5}$. Then $x=\frac{14}{5}$. Solution: $(\frac{14}{5},\frac{9}{5})$.

Q7. Solve: $x+3y=6$, $2x-y=3$.

Solution: From 1st: $x=6-3y$. Substitute in 2nd: $2(6-3y)-y=3\Rightarrow 12-6y-y=3\Rightarrow -7y=-9\Rightarrow y=\frac{9}{7}$. Then $x=6-3(\frac{9}{7})=\frac{15}{7}$. Solution: $(\frac{15}{7},\frac{9}{7})$.

Q8. Solve: $2x+5y=20$, $x-2y=-1$.

Solution: From 2nd: $x=2y-1$. Substitute in 1st: $2(2y-1)+5y=20\Rightarrow 4y-2+5y=20\Rightarrow 9y=22\Rightarrow y=\frac{22}{9}$. Then $x=2(\frac{22}{9})-1=\frac{35}{9}$. Solution: $(\frac{35}{9},\frac{22}{9})$.

Q9. Solve: $x+y=9$, $x-2y=4$.

Solution: From 1st: $x=9-y$. Substitute in 2nd: $9-y-2y=4\Rightarrow -3y=-5\Rightarrow y=\frac{5}{3}$. Then $x=\frac{22}{3}$. Solution: $(\frac{22}{3},\frac{5}{3})$.

Q10. Solve: $3x+y=10$, $2x-y=2$.

Solution: From 2nd: $y=2x-2$. Substitute in 1st: $3x+(2x-2)=10\Rightarrow 5x-2=10\Rightarrow 5x=12\Rightarrow x=\frac{12}{5}$. Then $y=2(\frac{12}{5})-2=\frac{14}{5}$. Solution: $(\frac{12}{5},\frac{14}{5})$.

FAQs (10 from NCERT)

1. Q: What is the substitution method? A: A method of solving equations by substituting one variable’s value into the other equation.

2. Q: What is the first step in substitution method? A: Solve one equation for one variable.

3. Q: What is the second step in substitution method? A: Substitute the value into the other equation.

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

5. Q: What is the fourth step in substitution method? A: Substitute back to find the first 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 substitution method useful? A: It gives exact solutions algebraically without graphing.

10. Q: Why is this exercise important?

Conclusion

Exercise 3.2 has 10 solved questions and 10 FAQs that strengthen your understanding of graphical solutions of linear equations in two variables. This builds the foundation for algebraic methods in Class 10 Maths.

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