Problem 56 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{y}{y^{2}+5 y+6}+\frac{4}{y^{2}-y-6}$$

Step-by-Step Solution

Answer

The simplified result of the given problem is \( \frac{y^{2}+y+12}{(y-3)(y+2)(y+3)} \).

Step 1: Factorize the denominators

Factorize both \(y^{2}+5y+6\) and \(y^{2}-y-6\). The factorized form of \(y^{2}+5y+6\) is \((y+2)(y+3)\) and \(y^{2}-y-6\) is \((y-3)(y+2)\). So, the fractions become \(\frac{y}{(y+2)(y+3)}\) and \(\frac{4}{(y-3)(y+2)}\) respectively.

Step 2: Find the Least Common Denominator (LCD)

To add fractions, we need to have a common denominator. For the denominators \((y+2)(y+3)\) and \((y-3)(y+2)\), the LCD is \((y-3)(y+2)(y+3)\).

Step 3: Convert into equivalent fractions with common denominator and add

We rewrite \(\frac{y}{(y+2)(y+3)}\) and \(\frac{4}{(y-3)(y+2)}\) as equivalent fractions with the LCD. So, the modified fractions are \(\frac{y(y-3)}{(y-3)(y+2)(y+3)}\) and \(\frac{4(y+3)}{(y-3)(y+2)(y+3)}\). We can now add the fractions: \(\frac{y(y-3)}{(y-3)(y+2)(y+3)} + \frac{4(y+3)}{(y-3)(y+2)(y+3)}\).

Step 4: Simplify the result

After adding, we get \(\frac{y^{2}-3y+4y+12}{(y-3)(y+2)(y+3)}\). On further simplification, we find the final result to be \(\frac{y^{2}+y+12}{(y-3)(y+2)(y+3)}\).