Subtract. Simplify, if possible.
-$$\frac { 7 x } { x ^ { 2 } - 9 } - \frac { x } { x - 3 }$$
A) $\frac { x ^ { 2 } + 4 x } { x ^ { 2 } - 9 }$

B) $\frac { - x ^ { 2 } + 10 x } { x ^ { 2 } - 9 }$

C) $\frac { x ^ { 2 } + 10 x } { x ^ { 2 } - 9 }$

D) $\frac { - x ^ { 2 } + 4 x } { x ^ { 2 } - 9 }$

Question

Subtract. Simplify, if possible.
-$$\frac { 7 x } { x ^ { 2 } - 9 } - \frac { x } { x - 3 }$$
A) $\frac { x ^ { 2 } + 4 x } { x ^ { 2 } - 9 }$

B) $\frac { - x ^ { 2 } + 10 x } { x ^ { 2 } - 9 }$

C) $\frac { x ^ { 2 } + 10 x } { x ^ { 2 } - 9 }$

D) $\frac { - x ^ { 2 } + 4 x } { x ^ { 2 } - 9 }$

Quizplus · Accepted Answer

To subtract these fractions, we need a common denominator. Factoring the first denominator, we get $x^2-9=(x+3)(x-3)$. Therefore, the common denominator is $(x+3)(x-3)$.

Using the common denominator, we can rewrite the expression as:

$$\frac{7x}{x^2-9}-\frac{x}{x-3}=\frac{7x}{(x+3)(x-3)}-\frac{x(x+3)}{(x+3)(x-3)}.$$

Combining the fractions, we get:

$$\frac{7x}{(x+3)(x-3)}-\frac{x(x+3)}{(x+3)(x-3)}=\frac{7x-x^2-3x}{(x+3)(x-3)}.$$

Simplifying the numerator, we have $7x-x^2-3x=-x^2+4x$, so the final answer is:

$$\frac{-x^2+4x}{(x+3)(x-3)}=\frac{-x(x-4)}{(x+3)(x-3)}=\frac{x(4-x)}{(x+3)(x-3)}.$$

This matches answer choice D.

Subtract Simplify, If Possible 7xx2−9−xx−3\frac { 7 x } { x ^ { 2 } - 9 } - \frac { x } { x - 3 }x2−97x​−x−3x​

Subtract Simplify, If Possible $\frac { 7 x } { x ^ { 2 } - 9 } - \frac { x } { x - 3 }$