Question

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

Answer 1

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

Given:

$\begin{array}{l}{2(a x-b y)+a-4 b=0} \\ {2(b x+a y)+b-4 a=0}\end{array}$

To find:

Value of x and y=?

Solution:

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

$\begin{array}{l}{2(a x-b y)+a-4 b=0} \\ {2(b x+a y)+b-4 a=0}\end{array}$

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

$\begin{array}{l}{a(2(a x-b y)+a-4 b)=0} \\ {b(2(b x+a y)+b-4 a)=0} \\ {((2 a a x-2 a b y)+a a-4 a b)=0} \\ {((2 b b x+2 b a y)+b b-4 a b)=0}\end{array}$

$\begin{array}{l}{\left(\left(2 a^{2} x-2 a b y\right)+a^{2}-4 a b\right)=0} \\ {\left(\left(2 b^{2} x+2 b a y\right)+b^{2}-4 a b\right)=0}\end{array}$

Subtracting the equations we get:

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

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

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

$\left(\left(2 b^{2} x+2 b a-2\right)+b^{2}-4 a b\right) ; \quad 2 a x=a ; \quad x=\frac{1}{2}$

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

Answer 2

Answer:

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

Step-by-step explanation:

It may be like this ,

2(ax-by)+a-4b=0 ---(1)

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

/* Multiply equation (1) by a and equation (2) by b , we get

2a²x-2aby+a²-4ab=0 ---(3)

2b²x+2aby+b²+4ab=0 ---(4)

/* Add (3) and (4) , we get

2x(a²+b²)+(a²+b²)=0

=> 2x(a²+b²)=-(a²+b²)

/* Divide both sides by (a²+b²), we get

=> 2x = -1

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

/* Put x value in equation (1), we get

$2a\times \frac{-1}{2}-2by+a-4b=0$

$\implies -a-2by+a-4b=0$

$\implies -2by-4b=0$

$\implies -2b(y+2)=0$

$\implies y+2=0$

$\implies y = -2$

Therefore,

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

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