Question

1. $\frac{y + 8}{y - 3}-\frac{10y + 30}{y^{2}-9}$

Explanation:

Step1: Factor the denominator

Since $y^{2}-9=(y + 3)(y - 3)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The given expression is $\frac{y + 8}{y - 3}-\frac{10y+30}{y^{2}-9}=\frac{y + 8}{y - 3}-\frac{10y + 30}{(y + 3)(y - 3)}$.

Step2: Find a common denominator

The common denominator of the two fractions is $(y + 3)(y - 3)$. Rewrite the first fraction: $\frac{y + 8}{y - 3}\times\frac{y + 3}{y + 3}=\frac{(y + 8)(y + 3)}{(y + 3)(y - 3)}$.

Step3: Expand the numerators

Expand $(y + 8)(y + 3)=y^{2}+3y+8y + 24=y^{2}+11y + 24$ and $10y + 30$. Then the expression becomes $\frac{y^{2}+11y + 24-(10y + 30)}{(y + 3)(y - 3)}$.

Step4: Simplify the numerator

$y^{2}+11y + 24-10y - 30=y^{2}+(11y-10y)+(24 - 30)=y^{2}+y - 6$.

Step5: Factor the numerator

Factor $y^{2}+y - 6=(y + 3)(y - 2)$ using the formula $x^{2}+(a + b)x+ab=(x + a)(x + b)$ where $a = 3$ and $b=-2$.

Step6: Simplify the fraction

$\frac{(y + 3)(y - 2)}{(y + 3)(y - 3)}=\frac{y - 2}{y - 3}$ for $y
eq - 3$.

Answer:

$\frac{y - 2}{y - 3},y
eq - 3$