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If $2 x - 3 y = 7$ and $(a + b) x - (a + b - 3) y = 4 a + b$ have infinite solutions. Then $(a, b) =$

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(- 5, - 1)

(- 5, 1)

(5, 1)

(5, - 1)

answer is A.

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Detailed Solution

Given, equations are,

2 x - 3 y = 7

and

(a + b) x - (a + b - 3) y = 4 a + b

For the standard system of equation,

a_{1} x + b_{1} y = c_{1}

a_{2} x + b_{2} y = c_{2}

Condition for infinite solutions,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \dots

… (1)
Here,

2 x - 3 y = 7

a_{1} = 2, b_{1} = - 3

and

c_{1} = 7

Also,

(a + b) x - (a + b - 3) y = 4 a + b,

a_{2} = a + b,

b_{2} = 3 - a - b

and

c_{2} = 4 a + b

Now for infinite many solutions,

\Rightarrow \frac{2}{a + b} = \frac{- 3}{3 - a - b} = \frac{7}{4 a + b} \dots

…. (2)
First two terms are,

\Rightarrow \frac{2}{a + b} = \frac{- 3}{3 - a - b}

\Rightarrow \frac{2 (3 - a - b)}{1} = \frac{- 3 (a + b)}{1}

\Rightarrow 6 - 2 a - 2 b = - 3 a - 3 b

\Rightarrow a + b = - 6

\Rightarrow a = - 6 - b \dots \dots

…. (3)
Similarly, as last two terms of eq. (2),

\Rightarrow \frac{- 3}{3 - a - b} = \frac{7}{4 a + b}

\Rightarrow \frac{- 3 (4 a + b)}{1} = \frac{7 (3 - a - b)}{1}

\Rightarrow - 12 a - 3 b = 21 - 7 a - 7 b

\Rightarrow 5 a - 4 b = - 21 \dots \dots \dots (4)

Putting the value of a in eq. (4),

5 a - 4 b = - 21

\Rightarrow 5 (- 6 - b) - 4 b = - 21 \Rightarrow - 30 - 5 b - 4 b = - 21 \Rightarrow 9 b = - 9

\Rightarrow b = - 1

Put the value of b in eq. (3),

a = - 6 - b

\Rightarrow a = - 6 - (- 1)

\Rightarrow a = - 5

Thus, the value of

(a, b) = (- 5, - 1)

.
Option 1 is correct.

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