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)

Respuesta :

Answer: Choice B

[tex]\frac{x}{(x+6)(x-2)} - \frac{3(x-2)}{(x+6)(x-2)}[/tex]

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

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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.

[tex]\frac{3}{x+6}[/tex] turns into [tex]\frac{3}{x+6}*\frac{x-2}{x-2} = \frac{3(x-2)}{(x+6)(x-2)}[/tex]

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So,

[tex]\frac{x}{x^2+4x-12} - \frac{3}{x+6}[/tex]

[tex]\frac{x}{(x+6)(x-2)} - \frac{3}{x+6}[/tex]

[tex]\frac{x}{(x+6)(x-2)} - \frac{3(x-2)}{(x+6)(x-2)}[/tex]

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.

Answer:

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

Step-by-step explanation: