Find all values of x satisfying the given conditions.
-$y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x + 3 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 7 x + 12 }$, and the sum of $y _ { 1 }$ and 2 times $y _ { 2 }$ is $y _ { 3 }$.
A) $\{ - 4 \}$
B) $\{ 0 \}$
C) $\{ 3 \}$
D) $\varnothing$

Question

Quizplus · Accepted Answer

Given that $y_1 + 2y_2 = y_3$, substituting the given expressions for $y_1$, $y_2$, and $y_3$ yields $\frac{1}{x+4} + \frac{2}{x+3} = \frac{-1}{x^2 + 7x + 12}$. Simplifying the right side of the equation, $x^2 + 7x + 12$ factors into $(x+4)(x+3)$, which are the denominators of the left side. However, the sign on the right side is negative, which cannot be achieved by adding positive fractions from the left side. Therefore, there is no solution.

Find All Values of X Satisfying the Given Conditions y1=1x+4,y2=1x+3,y3=−1x2+7x+12y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x + 3 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 7 x + 12 }y1​=x+41​,y2​=x+31​,y3​=x2+7x+12−1​

Find All Values of X Satisfying the Given Conditions $y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x + 3 } , y _ { 3 } = \frac { - 1 } { x ^ { 2 } + 7 x + 12 }$