Question

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

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

(a + b)(a + y) =  4ab

Answer 1

The given system of equation is

`(a - b)x + (a + b)y = 2a^2 - 2b^2`  .....(1)

(a + b)(a + y) =  4ab ....(ii)

`From equation (i), we get

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

`=> (a - b)x + (a - b) y - 2(a^2 - b^2 ) = 0` .....(iii)

From equation (ii), we get

`(a + b)x + (a + b)y - 4ab = 0` .....(iv)

Here

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

`a_2 = a + b, b_2 = a+ b, c_2 = -4ab`

By cross multiplication, we get

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

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

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

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

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

Now

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

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

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

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

Now

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

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

Hence `x = (2ab - a^2 + b^2)/b, y = ((a - b)(a^2 + b^2))/(b(a + b))` is the solution of the given system of equation

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

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

`=> y = (a^2d^2 - b^2c^2)/(a^4 - b^4)`