Question

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

`x(a - b + (ab)/(a -  b)) = y(a + b - (ab)/(a + b))`

`x + y = 2a^2`

Answer 1

The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

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

`x + y = 2a^2 = 0`

By cross multiplication method we get

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

`= 1/((a - b)+ (ab)/(a - b) - (-((a + b) - (ab)/(a + b))))`

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

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

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

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

`x/(((2a^4 + 2a^2b^2 + 2a^3b))/(a + b))= y/((2a^4 + 2a^2b^2 - 2a^3b)/(a - b))`

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

`x/((2a^4 + 2a^2b^2 + 2a^3b)/(a + b)) = y/((2a^4 + 2a^2b^2 - 2a^3b)/(a - b)) = 1/(((2a^3)/((a - b)(a + b)))`

Consider the following

`x/((2a^4 + 2a^2b^2 + 2a^3b)/(a + b)) = 1/(((2a^3)/((a - b)(a + b))))`

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

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

`x = (a^3 + ab^2 + a^2b - b^3 -ab^2 - a^2b)/a`

`x = (a^3 - b^3)/a`

And

`y/((2^4 + 2a^2b^2 -2a^3b)/(a - b))  =1 /((2a^3)/((a  -b)(a + b)))`

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

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

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

`y = (a^3 + b^3)/a`

Hence we get the value of `x = (a^3 - b^3)/a and y = (a^3 + b^3)/a`