Question

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

`2/x + 3/y = 13`

`5/4 - 4/y = -2`

where `x != 0 and y != 0`

Answer 1

The given system of equation is

``2/x + 3/y = 13`

`5/x - 4/y = -2 " where " x!= 0 and  y != 0`

Let `1/x = u` and `1/y = v`, Then the given system of equation becomes

2u + 3v = 13

5u - 4v = -2

here

`a_1 = 2, b_1 = 3, c_1 = -13`

`a_2 = 5, b_2 = -4, c_2 = 2`

By cross multiplication, we have

`=> u/(3xx2-(-13)xx(-4)) = (-v)/(2xx2-(-13)xx5) = 1/(2xx (-4)-3 xx 5)`

`=> u/(6 - 52) = (-v)/(4 + 65) = 1/(-8 - 15)`

`=> u/(-46) = (-v)/69 = 1/(-23)`

Now

`u/(-46) = 1/(-23)`

`=> u = (-46)/(-23) = 2`

And

`(-v)/69 =1/(-23)`

`=> v = (-69)/(-23) = 3`

Now

`x = 1/u = 1/2`

And

`y = 1/v = 1/3``

Hence `x = 1/2, y = 1/3` is the solution of the given system of equations.