what is the solution of $\frac{-8}{2y - 8}=\frac{5}{y + 4}-\frac{7y + 8}{y^{2}-16}$?\n$y=-4$\n$y=-2$\n$y =…

Explanation:

Step1: Factor the denominators

Factor (2y - 8=2(y - 4)) and (y^{2}-16=(y + 4)(y - 4)). The equation (\frac{-8}{2y - 8}=\frac{5}{y + 4}-\frac{7y + 8}{y^{2}-16}) becomes (\frac{-8}{2(y - 4)}=\frac{5}{y + 4}-\frac{7y + 8}{(y + 4)(y - 4)}).

Step2: Find the common - denominator

The common denominator of the fractions is (2(y + 4)(y - 4)). Multiply each term by the common denominator: (\frac{-8}{2(y - 4)}\times2(y + 4)(y - 4)=\frac{5}{y + 4}\times2(y + 4)(y - 4)-\frac{7y + 8}{(y + 4)(y - 4)}\times2(y + 4)(y - 4)). This simplifies to (-8(y + 4)=5\times2(y - 4)-2(7y + 8)).

Step3: Expand the expressions

Expand the right - hand side and left - hand side: (-8y-32 = 10y-40-(14y + 16)). (-8y-32=10y-40 - 14y-16).

Step4: Combine like terms

On the right - hand side, (10y-14y=-4y), and (-40-16=-56). So the equation is (-8y-32=-4y-56). Add (4y) to both sides: (-8y + 4y-32=-4y+4y-56), which gives (-4y-32=-56). Add 32 to both sides: (-4y-32 + 32=-56 + 32), so (-4y=-24).

Step5: Solve for y

Divide both sides by (-4): (y=\frac{-24}{-4}=6).

We need to check for extraneous solutions by substituting (y) into the original denominators. When (y = 6), (2y-8=2\times6 - 8 = 4\neq0), (y + 4=6 + 4 = 10\neq0), and (y^{2}-16=6^{2}-16=36 - 16 = 20\neq0).

Answer:

(y = 6)