Question

How do you solve `(10)/(x^2-2x)=4/x=5/(x-2)`?

Answer 1

First, we must change the denominators to become the same
( common denominator )

The first thing to do is to expand all the denominators, to fully see the components of the fractions.
`10/(x(x-2))=4/x=5/(x-2)`

If we multiply `4/x` by `(x-2)/(x-2)`, and `5/(x-2)` by `x/x`, then all three fractions will have the common denominator of `x(x-2)`:

`(10)/(x(x-2))=(4(x-2))/(x(x-2))=(5(x))/((x-2)(x))`

`10/(x^2-2x)=(4x-8)/(x^2-2x)=(5x)/(x^2-2x)`

multiply all fractions by `(x^2-2x)`

`cancel((x^2-2x))xx10/cancel((x^2-2x))=cancel((x^2-2x))xx(4x-8)/cancel((x^2-2x))=cancel((x^2-2x))xx(5x)/cancel((x^2-2x))`

We are left with

`10=(4x-8)=(5x)`