Solve the equation.
-$\frac{1}{y+2}-\frac{3}{y-2}=\frac{6}{y^{2}-4}$

Question

Quizplus · Accepted Answer

The equation simplifies to $\frac{1}{y+2}-\frac{3}{y-2}=\frac{6}{(y+2)(y-2)}$. Multiplying both sides by $(y+2)(y-2)$ to clear the denominators gives $y-2-3(y+2)=6$. Simplifying, we get $-2y=-12$, so $y=6$, which is not an option. However, there was a mistake in my calculation. Correctly simplifying the equation actually leads to a different result. Let's correct the process:Given: $\frac{1}{y+2}-\frac{3}{y-2}=\frac{6}{y^{2}-4}$First, notice that $y^2 - 4$ is the difference of squares, factoring to $(y+2)(y-2)$.The equation is already set with a common denominator on the right side, so the mistake was in the simplification and solving process. Let's correctly solve the equation without assuming an incorrect simplification:Multiplying through by the common denominator $(y+2)(y-2)$ to eliminate fractions:$(y+2)(y-2)\left(\frac{1}{y+2}-\frac{3}{y-2}\right)=(y+2)(y-2)\frac{6}{y^{2}-4}$This simplifies to:$y-2 - 3(y+2) = 6$Expanding:$y - 2 - 3y - 6 = 6$Combining like terms:$-2y - 8 = 6$$-2y = 14$$y = -7$So, the correct solution is $y = -7$, which corresponds to option B. My initial explanation contained a calculation error, leading to an incorrect conclusion.

Solve the Equation 1y+2−3y−2=6y2−4\frac{1}{y+2}-\frac{3}{y-2}=\frac{6}{y^{2}-4}y+21​−y−23​=y2−46​ A) 1\sqrt{1}1​ B) -7 C) 14 D) 7

Solve the Equation $\frac{1}{y+2}-\frac{3}{y-2}=\frac{6}{y^{2}-4}$ A) $\sqrt{1}$ B) -7
C) 14
D) 7