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Pair Of Linear Equations In Two Variables
EXERCISE 3.3
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

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) 2 1 and 3 2 3 3 x y y x    

Step-by-Step Solution

Key Concept: For a pair of linear equations in two variables, the <b>elimination method</b> involves making the coefficients of one variable equal (or opposite) in the two equations and then adding or subtracting the equations to eliminate that variable. The <b>substitution method</b> requires expressing one variable in terms of the other from one equation and substituting this expression into the second equation. Both methods lead to a single‑variable equation whose solution gives one variable; the other variable is then obtained by back‑substitution.
### (i) \(x + y = 5\) and \(2x - 3y = 4\)
Elimination
1. Multiply the first equation by 2: \(2x + 2y = 10\).
2. Subtract the second equation from this result:
\[(2x + 2y) - (2x - 3y) = 10 - 4\]
\[5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Substitute \(y\) in \(x + y = 5\):
\[x = 5 - \frac{6}{5} = \frac{19}{5}\].
Substitution
1. From the first equation, \(x = 5 - y\).
2. Substitute in the second equation:
\[2(5 - y) - 3y = 4 \Rightarrow 10 - 5y = 4 \Rightarrow 5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Hence \(x = 5 - \frac{6}{5} = \frac{19}{5}\).

### (ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
Elimination
1. Multiply the second equation by 2: \(4x - 4y = 4\).
2. Add to the first equation:
\[(3x + 4y) + (4x - 4y) = 10 + 4\]
\[7x = 14 \Rightarrow x = 2\].
3. Substitute \(x\) in \(2x - 2y = 2\):
\[4 - 2y = 2 \Rightarrow 2y = 2 \Rightarrow y = 1\].
Substitution
1. From the second equation, \(x - y = 1\) so \(x = y + 1\).
2. Substitute in the first equation:
\[3(y + 1) + 4y = 10 \Rightarrow 7y + 3 = 10 \Rightarrow 7y = 7 \Rightarrow y = 1\].
3. Hence \(x = 1 + 1 = 2\).

### (iii) \(3x - 5y = 4\) and \(9x = 2y + 7\)
Elimination
1. Rewrite the second equation as \(9x - 2y = 7\).
2. Multiply the first equation by 3: \(9x - 15y = 12\).
3. Subtract the second equation from this result:
\[(9x - 15y) - (9x - 2y) = 12 - 7\]
\[-13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Substitute \(y\) in \(3x - 5y = 4\):
\[3x + \frac{25}{13} = 4 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13}\].
Substitution
1. From the second equation, \(x = \frac{2y + 7}{9}\).
2. Substitute in the first equation:
\[3\left(\frac{2y + 7}{9}\right) - 5y = 4 \Rightarrow \frac{2y + 7}{3} - 5y = 4\].
3. Multiply by 3: \[2y + 7 - 15y = 12 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Hence \(x = \frac{2(-5/13) + 7}{9} = \frac{9}{13}\).

### (iv) \(2x + y = 1\) and \(3x - 2y = 3\)
Elimination
1. Multiply the first equation by 2: \(4x + 2y = 2\).
2. Add to the second equation:
\[(3x - 2y) + (4x + 2y) = 3 + 2\]
\[7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Substitute \(x\) in \(2x + y = 1\):
\[\frac{10}{7} + y = 1 \Rightarrow y = 1 - \frac{10}{7} = -\frac{3}{7}\].
Substitution
1. From the first equation, \(y = 1 - 2x\).
2. Substitute in the second equation:
\[3x - 2(1 - 2x) = 3 \Rightarrow 3x - 2 + 4x = 3 \Rightarrow 7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Hence \(y = 1 - 2\left(\frac{5}{7}\right) = -\frac{3}{7}\).

Correct Answer: {"(i)":{"x":"19/5","y":"6/5"},"(ii)":{"x":"2","y":"1"},"(iii)":{"x":"9/13","y":"-5/13"},"(iv)":{"x":"5/7","y":"-3/7"}}
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