Find all values of x satisfying the given conditions.
-$y _ { 1 } = \frac { 1 } { x + 5 } , y _ { 2 } = \frac { 3 } { x + 4 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 9 x + 20 }$, and $y _ { 1 } + y _ { 2 } = y _ { 3 }$
A) $\{ 4 \}$
B) $\{ 0 \}$
C) $\{ - 5 \}$
D) $\varnothing$

Question

Quizplus · Accepted Answer

Given $y _ { 1 } + y _ { 2 } = y _ { 3 }$, substituting the given expressions yields $\frac { 1 } { x + 5 } + \frac { 3 } { x + 4 } = \frac { - 1 } { x ^ { 2 } + 9 x + 20 }$. Simplifying the right side, $x^2 + 9x + 20$ factors into $(x + 5)(x + 4)$, making the equation $\frac { 1 } { x + 5 } + \frac { 3 } { x + 4 } = \frac { - 1 } { (x + 5)(x + 4) }$. Finding a common denominator and combining the fractions on the left side, we get $\frac { 4x + 13 } { (x + 5)(x + 4) } = \frac { - 1 } { (x + 5)(x + 4) }$. This equation suggests that $4x + 13 = -1$, which simplifies to $4x = -14$, leading to $x = -\frac{14}{4}$, which is not an option provided. Therefore, there is no solution from the given options.

Find All Values of X Satisfying the Given Conditions y1=1x+5,y2=3x+4,y3=−1x2+9x+20y _ { 1 } = \frac { 1 } { x + 5 } , y _ { 2 } = \frac { 3 } { x + 4 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 9 x + 20 }y1​=x+51​,y2​=x+43​,y3​=x2+9x+20−1​

Find All Values of X Satisfying the Given Conditions $y _ { 1 } = \frac { 1 } { x + 5 } , y _ { 2 } = \frac { 3 } { x + 4 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 9 x + 20 }$