Question

Darla found that the least common denominator needed to subtract \frac{x}{x^2+4x-12}-\frac{3}{x+6}is (x + 6)(x – 2). Which is the correct next step?
\frac{x−3}{x^2+3x−18}
x/(x+6)(x−2) – 3(x−2)/(x+6)(x−2)
x/(x+6)(x−2) – 3(x+6)/(x+6)(x−2)
x(x+6)(x−2)/(x+6)(x−2) – 3(x+6)(x−2)/(x+6)(x−2)

Answer 1

Answer: Choice B

$\frac{x}{(x+6)(x-2)} - \frac{3(x-2)}{(x+6)(x-2)}$

which is the same as x/((x+6)(x-2)) - 3(x-2)/((x+6)(x-2))

=====================================

Explanation:

The LCD is (x+6)(x-2) which is the factorization of x^2+4x-12, and that is the denominator of the first fraction. The first fraction has the LCD already. The second fraction does not. It has (x+6) but it is missing (x-2).

We multiply top and bottom of the second fraction by (x-2) to get the second fraction to have the LCD.

$\frac{3}{x+6}$ turns into $\frac{3}{x+6}*\frac{x-2}{x-2} = \frac{3(x-2)}{(x+6)(x-2)}$

------

So,

$\frac{x}{x^2+4x-12} - \frac{3}{x+6}$

$\frac{x}{(x+6)(x-2)} - \frac{3}{x+6}$

$\frac{x}{(x+6)(x-2)} - \frac{3(x-2)}{(x+6)(x-2)}$

This is the same as x/((x+6)(x-2)) - 3(x-2)/((x+6)(x-2))

Note the parenthesis around "(x+6)(x-2)"

Instead of x/(x+6)(x-2) you should write x/( (x+6)(x-2) ) to ensure that all of "(x+6)(x-2)" is in the denominator.