Question

How do you combine `\frac { x + 4} { 2x + 10} - \frac { 5} { x^2 - 25}` into one term?

Answer 1

See a solution process below:

Explanation:

First, factor the denominators of each fraction as:

`(x + 4)/(2(x + 5)) - 5/((x + 5)(x - 5))`

Next, multiply each fraction by the appropriate form of `1` to make the denominator `(2(x + 5)(x - 5))`

`[(x - 5)/(x - 5) xx (x + 4)/(2(x + 5))] - [2/2 xx 5/((x + 5)(x - 5))] =>`

`((x - 5)(x + 4))/(2(x + 5)(x - 5)) - (2 xx 5)/(2(x + 5)(x - 5)) =>`

`(x^2 - x - 20)/(2(x + 5)(x - 5)) - 10/(2(x + 5)(x - 5)) =>`

`(x^2 - x - 20)/(2(x + 5)(x - 5)) - 10/(2(x + 5)(x - 5))`

Then, we can subtract the numerators over the common denominator:

`(x^2 - x - 20 - 10)/(2(x + 5)(x - 5)) =>`

`(x^2 - x - 30)/(2(x + 5)(x - 5))`

Now, we can factor the numerator and simplify as:

`((x - 6)(x + 5))/(2(x + 5)(x - 5)) =>`

`((x - 6)color(red)(cancel(color(black)((x + 5)))))/(2color(red)(cancel(color(black)((x + 5))))(x - 5)) =>`

`(x - 6)/(2(x - 5))`

Or

`(x - 6)/(2x - 10)`

Answer 2

`color(magenta)((x-6)/(2(x-5))`

Explanation:

`(x+4)/(2x+10)-5/(x^2-25)`

`:.=(x+4)/(2(x+5))-5/((x+5)(x-5))`

`:.=((x+4)(x-5)-2(5))/(2(x+5)(x-5))`

`:.=((x+4)(cancel(x-5))^color(magenta)1)/(2(x+5)(cancel(x-5))^color(magenta)1)-cancel10^color(magenta)5/(cancel2^color(magenta)1(x+5)(x-5))`

`:.=(x+4)/(2(x+5))-5/((x+5)(x-5))`

`:.=((x+4)(x-5)-2(5))/(2(x+5)(x-5))`

`:.=(x^2-x-20-10)/(2(x^2-25))`

`:.=(x^2-x-30)/(2(x^2-25))`

`:.=((cancel(x+5))^color(magenta)1(x-6))/(2(cancel(x+5))^color(magenta)1(x-5))`

`:.color(magenta)(=(x-6)/(2(x-5))`