Question

How do you solve `(x)/(2x+7)=(x-5)/(x-1)`?

Answer 1

`x=-5` or `x=7`

Explanation:

Note : when you multiply something with a fraction, you multiply to the numerator.

First, let's simplify the LHS by multiplying `(x-5)/(x-1)` with `2x+7`

`(x-5*2x+7)/(x-1)`

`(2x^2-10x+7x-35)/(x-1)`

So we have

`x=(2x^2-3x-35)/(x-1)`

Then let's simplify the RHS by multiplying `x` with `x-1`.

`x/1=(2x^2-3x-35)/(x-1)` since `x=x/1`,

Then

`x*x-1=2x^2-3x-35`

Then `x^2-x=2x^2-3x-35`

`0=x^2-2x-35`

What we just did here is called cross multiplying.

Now, we have a quadratic equation.
We can factorise this to

`(x+5)(x-7) =0`

Set each factor equal to `0`, since if either `(x+5)` or `(x-7)` was 0, the equation would hold, because `0` multiplied by any number will always be 0.

`x+5=0`
`x=-5`

`x-7=0`
`x=7`

These are our two answers