First, write the value(s) that make the denominator(s) zero. Then solve the equation.
-$$\frac { x } { 2 x + 2 } = \frac { - 2 x } { 4 x + 4 } + \frac { 2 x - 3 } { x + 1 }$$
A) $\{ 3 \}$
B) $\left\{ \frac { 3 } { 2 } \right\}$
C) $\{ - 3 \}$
D) $\left\{ - \frac { 12 } { 5 } \right\}$

Question

Quizplus · Accepted Answer

The denominator of the first fraction is $2x+2$, which is zero when $x=-1$. The denominator of the second fraction is $4x+4$, which is zero when $x=-1$. The denominator of the third fraction is $x+1$, which is zero when $x=-1$. So, the only value that makes any of the denominators zero is $x=-1$.Substituting $x=-1$ into the equation, we get:$$\frac { -1 } { 2 (-1) + 2 } = \frac { - 2 (-1) } { 4 (-1) + 4 } + \frac { 2 (-1) - 3 } { -1 + 1 }$$$$\frac { -1 } { 0 } = \frac { 2 } { 0 } + \frac { -5 } { 0 }$$Since division by zero is undefined, the equation has no solution.

First, Write the Value(s) That Make the Denominator(s) Zero x2x+2=−2x4x+4+2x−3x+1\frac { x } { 2 x + 2 } = \frac { - 2 x } { 4 x + 4 } + \frac { 2 x - 3 } { x + 1 }2x+2x​=4x+4−2x​+x+12x−3​

First, Write the Value(s) That Make the Denominator(s) Zero $\frac { x } { 2 x + 2 } = \frac { - 2 x } { 4 x + 4 } + \frac { 2 x - 3 } { x + 1 }$