Question

Solve the following pairs of linear equations by elimination method:

$2(ax-by)+(a+4b)=0$ and $2(bx+ay)+(b-4a)=0$

• A. $x=6,y=3$
• B. $x=-1,y=3$
• C. $x=-\frac{1}{2},y=2$
• D. None of these

Answer 1

Answer: (C)

$x=-\frac{1}{2},y=2$

The given equations are

$2(ax-by)+(a+4b)=0$

$\Rightarrow 2ax-2by+a+4b=\mathrm{0..........}\left(i\right)$ and

$2(bx+ay)+b-4a=0$

$\Rightarrow 2bx+2ay+b-4a=\mathrm{0.........}\left(ii\right)$

Multiplying $\left(i\right)$ by $b$ and $\left(ii\right)$ by $a$ and subtracting $\left(ii\right)\ast a$ from $\left(i\right)\ast b$, we get

$-y(2b^2+2a^2)+(4b^2+4a^2)=0\Rightarrow y=\frac{2(2b^2+2a^2)}{(2b^2+2a^2)}=2.$

Substituting $y=2$ in $\left(i\right)$,

$2ax=-a\Rightarrow x=-\frac{1}{2}.\therefore (x,y)=(-\frac{1}{2},2).$

Hence option $C$