Question

How do you solve `2/(x(x+2))+3/x=4/(x-2)`?

Answer 1

We can solve this by multiplying everything by the least common multiple `x(x+2)(x-2)`

Explanation:

Note: We have to reject the solutions `x=0,x=2,x=-2`, as they would make one or more of the fractions undetermined.

The factors not in the denominators will not be canceled and end up in the numerators:
`->(2*cancelx(cancel(x+2))(x-2))/(cancelx(cancel(x+2)))+(3*cancelx(x+2)(x-2))/cancelx=(4*x(x+2)(cancel(x-2)))/cancel(x-2)`

`->2(x-2)+3(x+2)(x-2)=4x(x+2)`

Work out the parentheses and put everything to one side:
`->2x-4+3x^2-12=4x^2+8x`

`->-x^2-6x-16 =0->x^2+6x+16=0`

The Discriminant `D=b^2-4ac=6^2-4*1*16=-28<0`

Since `D<0`, there are no real solutions.