Question

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

Answer 1

2  and \bold{-\frac{1}{2}}−21 is the value of x and y.

Given:

\begin{lgathered}\begin{array}{l}{2(a x-b y)+a-4 b=0} \\ {2(b x+a y)+b-4 a=0}\end{array}\end{lgathered}2(ax−by)+a−4b=02(bx+ay)+b−4a=0

To find:

Value of x and y=?

Solution:

To find the value of x and y let us solve the equation first that is  

\begin{lgathered}\begin{array}{l}{2(a x-b y)+a-4 b=0} \\ {2(b x+a y)+b-4 a=0}\end{array}\end{lgathered}2(ax−by)+a−4b=02(bx+ay)+b−4a=0

First multiplying (a) to the first equation and (b) to the second equation, we get

\begin{lgathered}\begin{array}{l}{a(2(a x-b y)+a-4 b)=0} \\ {b(2(b x+a y)+b-4 a)=0} \\ {((2 a a x-2 a b y)+a a-4 a b)=0} \\ {((2 b b x+2 b a y)+b b-4 a b)=0}\end{array}\end{lgathered}a(2(ax−by)+a−4b)=0b(2(bx+ay)+b−4a)=0((2aax−2aby)+aa−4ab)=0((2bbx+2bay)+bb−4ab)=0

\begin{lgathered}\begin{array}{l}{\left(\left(2 a^{2} x-2 a b y\right)+a^{2}-4 a b\right)=0} \\ {\left(\left(2 b^{2} x+2 b a y\right)+b^{2}-4 a b\right)=0}\end{array}\end{lgathered}((2a2x−2aby)+a2−4ab)=0((2b2x+2bay)+b2−4ab)=0

Subtracting the equations we get:

-2 y\left(a^{2}+b^{2}\right)=4\left(b^{2}+a^{2}\right)−2y(a2+b2)=4(b2+a2)

Solving the above equation we get the value of y = -2

Putting the value of y in\left(\left(2 b^{2} x+2 b a y\right)+b^{2}-4 a b\right)((2b2x+2bay)+b2−4ab) we get the value of x as  

\left(\left(2 b^{2} x+2 b a-2\right)+b^{2}-4 a b\right) ; \quad 2 a x=a ; \quad x=\frac{1}{2}((2b2x+2ba−2)+b2−4ab);2ax=a;x=21

Therefore, the value of \bold{x=\frac{1}{2}, y=-2.}x=21,y=−2.