First, write the value(s) that make the denominator(s) zero. Then solve the equation.
-$\frac { 7 } { x + 1 } + \frac { 3 } { x - 1 } = \frac { 6 } { ( x + 1 ) ( x - 1 ) }$
A) $x \neq - 1 ; \{ 1 \}$
B) $x \neq - 1,1 ; \{ 2 \}$
C) No restrictions; $\{ 1 \}$
D) $x \neq - 1,1 ; \varnothing$

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The Answer of First, write the value(s) that make the denominator(s) zero. Then solve the equation.
-$\frac { 7 } { x + 1 } + \frac { 3 } { x - 1 } = \frac { 6 } { ( x + 1 ) ( x - 1 ) }$
A) $x \neq - 1 ; \{ 1 \}$
B) $x \neq - 1,1 ; \{ 2 \}$
C) No restrictions; $\{ 1 \}$
D) $x \neq - 1,1 ; \varnothing$

First, Write the Value(s) That Make the Denominator(s) Zero 7x+1+3x−1=6(x+1)(x−1)\frac { 7 } { x + 1 } + \frac { 3 } { x - 1 } = \frac { 6 } { ( x + 1 ) ( x - 1 ) }x+17​+x−13​=(x+1)(x−1)6​

First, Write the Value(s) That Make the Denominator(s) Zero $\frac { 7 } { x + 1 } + \frac { 3 } { x - 1 } = \frac { 6 } { ( x + 1 ) ( x - 1 ) }$