#### Question

Solve each of the following systems of equations by the method of cross-multiplication :

2(ax – by) + a + 4b = 0

2(bx + ay) + b – 4a = 0

#### Solution

The given system of equation may be written as

2(ax – by) + a + 4b = 0

2(bx + ay) + b – 4a = 0

here

`a_1 = 2a, b_1 = -2b, c_1 = a+ 4b`

`a_2 = 2b, b_2 = 2a, c_2 = b - 4a`

By cross multiplication, we have

`=> x/(((-2b)(b - 4a) - (2a)(a + 4b))) = (-y)/((2b)(b - 4a) - (2a) (a + 4a) ) = 1/(4a^2 + 4b^2)`

`=> x/(-2b^2 + 8ab - 2a^2 - 8ab) = (-y)/(2ab - 8a^2 - 2ab - 8b^2) = 1/(4a^2 + 4b^2)`

`=> x/(-2a^2 - 2b^2) = (-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)`

Now

`x/(-2a^2 - 2b^2) = 1/(4a^2 + 4b^2)`

`=> x = (-2a^2 - 2b^2)/(4a^2 + 4b^2)`

`= (-2(a^2 - b^2))/(4(a^2 + b^2))`

`= (-1)/2`

And

`(-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)`

`=> -y = (-8a^2 - 8b^2)/(4a^2 + 4b^2)`

`=> -y = (-8(a^2 - b^2))/(4(a^2 + b^2))`

`=> -y = (-8)/4`

=> y = 2

Hence x = (-1)/2, y = 2 is the solution of the given system of the equations.