First, write the value(s) that make the denominator(s) zero. Then solve the equation.
-$y _ { 1 } = \frac { 1 } { x + 7 } , y _ { 2 } = \frac { 4 } { x + 5 } , y _ { 3 } = \frac { - 2 } { x ^ { 2 } + 12 x + 35 }$, and $y _ { 1 } + y _ { 2 } = y _ { 3 }$
A) $\{ 0 \}$
B) $\{ - 7 \}$
C) $\{ 5 \}$
D) $\varnothing$

Question

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The Answer of First, write the value(s) that make the denominator(s) zero. Then solve the equation.
-$y _ { 1 } = \frac { 1 } { x + 7 } , y _ { 2 } = \frac { 4 } { x + 5 } , y _ { 3 } = \frac { - 2 } { x ^ { 2 } + 12 x + 35 }$, and $y _ { 1 } + y _ { 2 } = y _ { 3 }$
A) $\{ 0 \}$
B) $\{ - 7 \}$
C) $\{ 5 \}$
D) $\varnothing$

First, Write the Value(s) That Make the Denominator(s) Zero y1=1x+7,y2=4x+5,y3=−2x2+12x+35y _ { 1 } = \frac { 1 } { x + 7 } , y _ { 2 } = \frac { 4 } { x + 5 } , y _ { 3 } = \frac { - 2 } { x ^ { 2 } + 12 x + 35 }y1​=x+71​,y2​=x+54​,y3​=x2+12x+35−2​

First, Write the Value(s) That Make the Denominator(s) Zero $y _ { 1 } = \frac { 1 } { x + 7 } , y _ { 2 } = \frac { 4 } { x + 5 } , y _ { 3 } = \frac { - 2 } { x ^ { 2 } + 12 x + 35 }$