Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$$
A) $\left\{ \frac { 2 } { 3 } , - 6 \right\}$

B) $\left\{ - \frac { 2 } { 3 } , 6 \right\}$

C) $\left\{ - \frac { 3 } { 2 } , - 6 \right\}$

D) $\{ 2,6 \}$

Question

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$$
A) $\left\{ \frac { 2 } { 3 } , - 6 \right\}$

B) $\left\{ - \frac { 2 } { 3 } , 6 \right\}$

C) $\left\{ - \frac { 3 } { 2 } , - 6 \right\}$

D) $\{ 2,6 \}$

Quizplus · Accepted Answer

The Answer of Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$$
A) $\left\{ \frac { 2 } { 3 } , - 6 \right\}$

B) $\left\{ - \frac { 2 } { 3 } , 6 \right\}$

C) $\left\{ - \frac { 3 } { 2 } , - 6 \right\}$

D) $\{ 2,6 \}$

Decide What Values of the Variable Cannot Possibly Be Solutions 8x−2+8x+2=3\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3x−28​+x+28​=3

Decide What Values of the Variable Cannot Possibly Be Solutions $\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$