Question

For what value of $a$ and $b$, the following system of equations have an infinite number of solutions.
$2x+3y=7$; $(a-b)x+(a+b)y=3a+b-2$

• A. $a=-9,b=16$
• B. $a=7,b=-13$
• C. $a=-6,b=2$
• D. $a=5,b=1$

Answer 1

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

Given equations are,

$2x+3y=7$
and $(a-b)x+(a+b)y=3a+b-2$
For infinite number of solutions:
$\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$
$\Rightarrow \frac{2}{a-b}=\frac{3}{a+b}$
$\Rightarrow 2a+2b=3a-3b$
$\Rightarrow a=5b$ ......(i)
$\frac{3}{a+b}=\frac{7}{3a+b-2}$
$\Rightarrow 9a+3b-6=7a+7b$
$\Rightarrow 2a-4b=6$ .....(ii)
Substitute $a=5b$ in equation (ii), we get
$2\left(5b\right)-4b=6$
$\Rightarrow 6b=6$
$\Rightarrow b=1$
So, when $b=1,a=5\times 1=5$
Hence, $a=5,b=1$