Subtract. Simplify, if possible.
-$$\frac { b } { b - c } - \frac { b } { b + c }$$
A) $\frac { 2 b ^ { 2 } - 2 b c } { ( b - c ) ^ { 2 } }$

B) $- \frac { b } { c }$

C) $\frac { 2 b c } { ( b - c ) ( b + c ) }$

D) 0

Question

Subtract. Simplify, if possible.
-$$\frac { b } { b - c } - \frac { b } { b + c }$$
A) $\frac { 2 b ^ { 2 } - 2 b c } { ( b - c ) ^ { 2 } }$

B) $- \frac { b } { c }$

C) $\frac { 2 b c } { ( b - c ) ( b + c ) }$

D) 0

Quizplus · Accepted Answer

To subtract these fractions, we need a common denominator. In this case, the common denominator is (b-c)(b+c). So we get:

$$\frac{b(b+c)}{(b-c)(b+c)} - \frac{b(b-c)}{(b-c)(b+c)}$$

Simplifying further gives:

$$\frac{2bc}{(b-c)(b+c)} = \frac{2bc}{b^2-c^2}$$

This is equivalent to choice C.

Subtract Simplify, If Possible bb−c−bb+c\frac { b } { b - c } - \frac { b } { b + c }b−cb​−b+cb​