Question

How do you add `\frac { x } { x + 3} + \frac { 6x - 2} { x - 3}`?

Answer 1

`x/(x+3)+(6x-2)/(x-3)=(7x^2+13x-6)/(x^2-9)`

Explanation:

Step 1: Find the least common multiple of all denominators

`x+3` and `x-3` have no shared factors, so their least common multiple is their product: `(x+3)(x-3)`

Step 2: Multiply each term by an appropriate `1` to give them common denominators

`x/(x+3) = x/(x+3) * (x-3)/(x-3)= (x(x-3))/((x+3)(x-3))`

`(6x-2)/(x-3) = (6x-2)/(x-3) * (x+3)/(x+3) = ((6x-2)(x+3))/((x+3)(x-3))`

Step 3: Add the numerators of the new fractions

`x/(x+3)+(6x-2)/(x-3) = (x(x-3))/((x+3)(x-3))+((6x-2)(x+3))/((x+3)(x-3))`

`=(x(x-3)+(6x-2)(x+3))/((x+3)(x-3))`

Step 4: Simplify

`(x(x-3)+(6x-2)(x+3))/((x+3)(x-3)) = ((x^2-3x)+(6x^2+16x-6))/(x^2-9)`

`=(7x^2+13x-6)/(x^2-9)`