Find all values of x satisfying the given conditions.
-$y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x - 4 } , y _ { 3 } = \frac { 1 } { x ^ { 2 } - 16 }$, and the difference between 4 times $y _ { 1 }$ and 2 times $y _ { 2 }$ is the product of 6 and $y 3$.
A) $\{ 30 \}$
B) $\{ 15 \}$
C) $\{ - 15 \}$
D) $\{ 3 \sqrt { 2 } \}$

Question

Quizplus · Accepted Answer

The given condition can be translated into the equation: $4y_1 - 2y_2 = 6y_3$. Substituting the given expressions for $y_1$, $y_2$, and $y_3$ yields: $4\left(\frac{1}{x+4}\right) - 2\left(\frac{1}{x-4}\right) = 6\left(\frac{1}{x^2-16}\right)$. Simplifying and solving for $x$ gives $x = 15$.

Find All Values of X Satisfying the Given Conditions y1=1x+4,y2=1x−4,y3=1x2−16y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x - 4 } , y _ { 3 } = \frac { 1 } { x ^ { 2 } - 16 }y1​=x+41​,y2​=x−41​,y3​=x2−161​

Find All Values of X Satisfying the Given Conditions $y _ { 1 } = \frac { 1 } { x + 4 } , y _ { 2 } = \frac { 1 } { x - 4 } , y _ { 3 } = \frac { 1 } { x ^ { 2 } - 16 }$