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

Question 363

Multiple Choice

Decide what values of the variable cannot possibly be solutions for the equation.
- 8x−2+8x+2=3\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3


A) {23,−6}\left\{ \frac { 2 } { 3 } , - 6 \right\}

B) {−23,6}\left\{ - \frac { 2 } { 3 } , 6 \right\}

C) {−32,−6}\left\{ - \frac { 3 } { 2 } , - 6 \right\}

D) {2,6}\{ 2,6 \}

Correct Answer:

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