Question

For which values of 'a' and 'b' does the following pair of linear equations have an infinite number of solutions
$2x+3y=7,(a-b)x+(a+b)y=3a+b-2$

• A. $a=5,b=1$
• B. $a=4,b=2$
• C. $a=1,b=5$
• D. $a=2,b=4$

Answer 1

The correct option is A $a=5,b=1$

$2x+3y=7$
$(a-b)x+(a+b)y=3a+b-2$
$\because$ It has infinitely many solutions.
$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\underset{\left(i\right)}{\frac{2}{a-b}}=\underset{\left(ii\right)}{\frac{3}{a+b}}=\underset{\left(iii\right)}{\frac{7}{3a+b-2}}$
From $\left(i\right)$ and $\left(ii\right)$
$\frac{3}{a+b}=\frac{7}{3a+b-2}$
$7a+7b=9a+3b-6$
$\Rightarrow 2a-4b=6$
$\Rightarrow a-2b=3$
$\Rightarrow 5b-2b=3$
$\Rightarrow 3b=3\Rightarrow b=1$
$\therefore a=5\times 1=5$.