Question

Solve the following system of linear equations:
$2(ax-by)+(a+4b)=0$
$2(bx+ay)+(b-4a)=0$

• A. $x=-\frac{1}{3},y=2$
• B. $x=\frac{1}{2},y=2$
• C. $x=-\frac{1}{2},y=2$
• D. $x=-\frac{1}{2},y=3$

Answer 1

The correct option is C $x=-\frac{1}{2},y=2$

$2(ax-by)+(a+4b)=0\times b$ __(1) [Assuming a and b are non zero constant]

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

(1) $2(abx-b^{2}y)+(ab+4b^2)=0$

(2) $2(abx+a^{2}y)+(ab-4a^2)=0$

(2) - (1) $2(a^2+b^2)y-4(a^2+b^2)=0$

$\Rightarrow 2y-4=0$

$\Rightarrow y=2$

substituting $y=2$ in (1)

$ab(1+2x)=0$

$\Rightarrow x=\frac{-1}{2}$

$\therefore$ The solution for the simultaneous equation is :

$x=-1/2,y=2$