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

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Answer 1 · 2016-12-29T00:08:29

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

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

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

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

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