Question

Solve the following system of equations by the method of cross-multiplication:

`a^2/x - b^2/y = 0`

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

Answer 1

Taking `1/x = u and 1/y = v`

Then the given system of equations become

`a^2u - b^2v = 0`

`a^2bu + b^2av - (a + b) = 0`

Here

`a_1 = a^2, b_1 = -b^2, c_1 = 0`

`a_2 = a^2b, b_2 = b^2a, c_2 = -(a + b)`

By cross multiplication, we have

`=> u/(b^2(a + b)-0xxb^2a) = (-v)/(-a^2(a + b)-0xxa^2b) = 1/(a^3b^2 + a^2b^3)`

`=> u/(b^2(a + b)) = v/(a^2(a + b)) = 1/(a^2b^2(a + b))`

Now

`u/(b^2(a + b)) = 1/(a^2b^2(a + b))`

`=> u = (b^2(a + b))/(a^2b^2 (a + b))`

`=> u = 1/a^2`

And

`v/(a^2(a + b)) = 1/(a^2b^2(a + b))`

`=> v = (a^2 (a + b))/(a^2b^2 (a + b))`

`=> v = 1/b^2`

Now

`x = 1/u = a^2`

And

`y = 1/v = b^2`

Hence `x = a^2, y = b^2` is the solution of the given system of equations.