Question

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

bx + cy  = a + b

`ax (1/(a - b) - 1/(a + b)) + cy(1/(b -a) - 1/(b + a)) = (2a)/(a + b)`

Answer 1

The given system of equation is

bx + cy  = a + b  .....(i)

`ax (1/(a - b) - 1/(a + b)) + cy(1/(b -a) - 1/(b + a)) = (2a)/(a + b)`  ......(ii)

From equation (ii), we get

bx + cy - (a + b) = 0 .....(iii)

From equation (ii), we get

`ax[(a + b-(a-b))/((a - b)(a + b))] + cy((b + a - (b -a))/((b-a)(b + a))) - (2a)/(a + b) = 0`

`=> ax [(a + b - a + b)/((a - b)(a + b))] + cy((b + a - b + a)/((b -a)(b+a))) - (2a)/(a + b) = 0`

`=> ax[(2b)/(a - b)(a + b)] + cy((2a)/((b -a)(b+a))) - (2a)/(a + b) = 0`

`=> x [(2ab)/((a - b)(a + b))] + y((2ac)/((a - b)(a + b))) - (2a)/(a +b) = 0`

`=> 1/(a + b)[(2abx)/(a - b) - (2acy)/(a -b) - 2a] = 0`

`=> (2abx - 2acy - 2a(a - b))/(a - b) = 0`

`=> 2abx -  2acy - 2a(a - b) = 0` .....(iv)

From equation (i) and equation (ii), we get

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

`a_2 = 2ab,b_2 = -2ac, c_2 = -2a(a - b)`

By cross multiplication, we get

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

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

`=> x/(-4a^2c) = (-y)/(4ab^2) = (-1)/(4abc)`

Now

`x/(-4a^2c) = (-1)/(4abc)`

`=> x = (4a^2c)/(4abc) = a/b`

And

`(-y).(4ab^2) = (-1)/(4abc)`

`=> y = (4ab^2)/(4abc) = b/c`

Hence `x = a/b, y = b/c` isthe soluiton of the given system of the equations.