NCERT Solutions
Class 10 - Mathematics - Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - Exercise 3.3
Question 1

Question. 1

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) \(x + y = 5\) and \(2x - 3y = 4\)

(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)

(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)

(iv) \(\dfrac{x}{2} + \dfrac{2y}{3} = -1\) and \(x - \dfrac{y}{3} = 3\)

Answer:

(i) \(x + y = 5\), \(2x - 3y = 4\)

From \(x + y = 5\), express \(x\) as \(x = 5 - y\). Substitute in \(2x - 3y = 4\):

\[2(5 - y) - 3y = 4 \Rightarrow 10 - 2y - 3y = 4 \Rightarrow 10 - 5y = 4.\]

\[-5y = -6 \Rightarrow y = \dfrac{6}{5}.\]

Then \(x = 5 - y = 5 - \dfrac{6}{5} = \dfrac{25 - 6}{5} = \dfrac{19}{5}.\)

Solution: \(x = \dfrac{19}{5},\ y = \dfrac{6}{5}.\)

(ii) \(3x + 4y = 10\), \(2x - 2y = 2\)

From \(2x - 2y = 2\), divide by 2:

\[x - y = 1 \Rightarrow x = y + 1.\]

Substitute in \(3x + 4y = 10\):

\[3(y + 1) + 4y = 10 \Rightarrow 3y + 3 + 4y = 10 \Rightarrow 7y + 3 = 10.\]

\[7y = 7 \Rightarrow y = 1, \quad x = y + 1 = 2.\]

Solution: \(x = 2,\ y = 1.\)

(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)

Rewrite the equations:

\[3x - 5y = 4, \quad 9x - 2y = 7.\]

Multiply the first equation by 2:

\[6x - 10y = 8.\]

Multiply the second equation by 5:

\[45x - 10y = 35.\]

Subtract the first (multiplied) equation from the second:

\[(45x - 10y) - (6x - 10y) = 35 - 8 \Rightarrow 39x = 27.\]

\[x = \dfrac{27}{39} = \dfrac{9}{13}.\]

Substitute in \(3x - 5y = 4\):

\[3 \cdot \dfrac{9}{13} - 5y = 4 \Rightarrow \dfrac{27}{13} - 5y = 4.\]

\[-5y = 4 - \dfrac{27}{13} = \dfrac{52 - 27}{13} = \dfrac{25}{13} \Rightarrow y = -\dfrac{25}{65} = -\dfrac{5}{13}.\]

Solution: \(x = \dfrac{9}{13},\ y = -\dfrac{5}{13}.\)

(iv) \(\dfrac{x}{2} + \dfrac{2y}{3} = -1\), \(x - \dfrac{y}{3} = 3\)

Clear denominators by multiplying each equation by 6:

First equation:

\[6\left(\dfrac{x}{2} + \dfrac{2y}{3}\right) = 6(-1) \Rightarrow 3x + 4y = -6.\]

Second equation:

\[6\left(x - \dfrac{y}{3}\right) = 6 \cdot 3 \Rightarrow 6x - 2y = 18.\]

Divide the second equation by 2 to simplify:

\[3x - y = 9.\]

Now solve the pair \(3x + 4y = -6\) and \(3x - y = 9\) by elimination. Subtract the second from the first:

\[(3x + 4y) - (3x - y) = -6 - 9 \Rightarrow 5y = -15 \Rightarrow y = -3.\]

Substitute in \(3x - y = 9\):

\[3x - (-3) = 9 \Rightarrow 3x + 3 = 9 \Rightarrow 3x = 6 \Rightarrow x = 2.\]

Solution: \(x = 2,\ y = -3.\)

Detailed Answer with Explanation:

Clear decimals or fractions, then use substitution or elimination to remove one variable. Solve the simplified equation, back-substitute, and confirm the ordered pair satisfies both equations.

NCERT Solutions – Chapter-wise Questions & Answers