Question

Find the value of a and b for which the given system of linear equation has an infinite number of solutions.

$2x+3y=7$ and $(a-b)x+(a+b)y=3a+b-2$

• A. $a=2,b=3$
• B. $a=5,b=1$
• C. $a=-5,b=3$
• D. $a=-1,b=2$

Answer 1

Answer: (B)

$a=5,b=1$

Comparing $2x+3y=7$ with $a_{1}x+b_{1}y+c_1=0$,

we get,

$a_1=2,b_1=3,c_1=-7$

Comparing $(a-b)x+(a+b)y=3a+b-2$ with $a_{2}x+b_{2}y+c_2=0$,

we get,

$a_2=(a-b),b_2=(a+b),c_2=-(3a+b-2)$

Given system of equation have infinite solutions:
$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

From $\frac{a_1}{a_2}=\frac{b_1}{b_2}$

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

$\Rightarrow 2a+2b=3a-3b$

$\Rightarrow a-5b=0.........\left(I\right)$

Now from $\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

$\Rightarrow 9a+3b-6=7a+7b$

$\Rightarrow 9a-7a+3b-7b=6$

$\Rightarrow 2a-4b=6......\left(II\right)$

Now, for solving eq $I$ and eq $II$

$a-5b=0$

$2a-4b=6$

Multiplying eq $I$ by $2$

$2a-10b=0$

$2a-4b=6$

on subtraction we get

$\Rightarrow -6b=-6$

$\Rightarrow b=1$

Now, substituting $b$ value in eq $I$

$\Rightarrow 2a-10\left(1\right)=0$

$2a-10=0$

$2a=10$

$a=\frac{10}{2}$

$a=5$

$\Rightarrow a=5,b=1$