Question

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

`x/a + y/b = 2`

`ax - by = a^2 - b^2`

Answer 1

The system of the given equations may be written as

`1/a x xx + 1/b xx y - 2 = 0`

`ax - by + b^2 - a^2 = 0`

here

`a_1 = 1/a, b_1 = 1/b, c_1 = -2`

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

By cross multiplication, we get

`=> x/(1/b xx (b^2 - a^2) - (-2) xx (-b)) = (-y)/(1/a xx (b^2 - a^2) - (-2) xx a) = 1/((-bxx1)/a - (a xx1)/b)`

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

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

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

Now

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

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

And

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

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

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

=> y = b

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