Question

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

`b/a x + a/b y - (a^2 + b^2) = 0`

x + y - 2ab = 0

Answer 1

The given system of the equation may be written as

`b/a x + a/b y - (a^2 + b^2) = 0``

x + y - 2ab = 0

Here

`a_1 = b/a, b_1 =- a/b,c_1 = -(a^2 + b^2)`

`a_2 = 1, b_2 = 1, c_2 = -2ab`

By cross multiplication, we have

`x/(-2ab  xx a/b + a^2 + b^2) = (-y)/(-2ab xx a/b + a^2 + b^2) = 1/(b/a - a/b)`

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

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

Now

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

`=> x = b^2 - a^2 aa (ab)/(b^2 - a^2)`

=> x = ab

And

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

`=> -y = b^2 + a^2 xx (ab)/(b^2 - a^2)`

`=> -y =-(b^2 - a^2) xx (ab)/(b^2 - a^2)``

=> -y = -ab

=> y = ab

Hence, x = ab,y = ab is the solution of the given system of equations.