Find all values of x satisfying the given conditions.
-$y _ { 1 } = \frac { 1 } { x + 6 } , y _ { 2 } = \frac { 2 } { x + 3 } , y _ { 3 } = \frac { - 3 } { x ^ { 2 } + 9 x + 18 }$, and $y _ { 1 } + y _ { 2 } = y _ { 3 }$
A) $\{ 0 \}$
B) $\{ - 6 \}$
C) $\{ 3 \}$
D) $\varnothing$

Question

Quizplus · Accepted Answer

Given $y_1 + y_2 = y_3$, substituting the given expressions yields $\frac{1}{x+6} + \frac{2}{x+3} = \frac{-3}{x^2 + 9x + 18}$. Simplifying the right side, $x^2 + 9x + 18$ can be factored as $(x+6)(x+3)$, making the equation $\frac{1}{x+6} + \frac{2}{x+3} = \frac{-3}{(x+6)(x+3)}$. Multiplying both sides by $(x+6)(x+3)$ to clear the denominators, we get $1(x+3) + 2(x+6) = -3$, which simplifies to $3x + 15 = -3$, or $3x = -18$, giving $x = -6$. However, substituting $x = -6$ back into the original expressions for $y_1, y_2, y_3$ results in division by zero, which is undefined. Therefore, there are no values of $x$ that satisfy the given conditions.

Find All Values of X Satisfying the Given Conditions y1=1x+6,y2=2x+3,y3=−3x2+9x+18y _ { 1 } = \frac { 1 } { x + 6 } , y _ { 2 } = \frac { 2 } { x + 3 } , y _ { 3 } = \frac { - 3 } { x ^ { 2 } + 9 x + 18 }y1​=x+61​,y2​=x+32​,y3​=x2+9x+18−3​

Find All Values of X Satisfying the Given Conditions $y _ { 1 } = \frac { 1 } { x + 6 } , y _ { 2 } = \frac { 2 } { x + 3 } , y _ { 3 } = \frac { - 3 } { x ^ { 2 } + 9 x + 18 }$