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. $a=1$ and $b=7$
• B. $a=13$ and $b=6$
• C. $a=2$ and $b=10$
• D. $a=5$ and $b=1$

Answer 1

The correct option is D $a=5$ and $b=1$

Given Pair of equations are:
$2x+3y-7=0$ .......(i)
$(a-b)x+(a+b)y-(3a+b-2)=0$ ........(ii)
For infinite number of solutions, we have
$\frac{a-b}{2}=\frac{a+b}{3}=\frac{3a+b-2}{7}$
For first and second, we have
$\frac{a-b}{2}=\frac{a+b}{3}$ or $3a-3b=2a+2b$
$a=5b$ .......(i)
From second and third, we have
$\frac{a+b}{3}=\frac{3a+b-2}{7}$
$7a+7b=9a+3b-6$

$\Rightarrow 4b=2a-6$
$\Rightarrow 2b=a-3$ .......(ii)
From (i) and (ii), eliminating $a$,
$2b=5b-3$
$\Rightarrow 2b-5b=-3$
$\Rightarrow -3b=-3$
$\Rightarrow b=1$
Substituting $b=1$ in (i), we get $a=5$