Answer
$\frac{2}{g + 1}$
Work Step by Step
To add or subtract rational expressions, we need to make sure that the expressions have the same denominator.
We need to rewrite the two expressions with the same denominator, so we need to find the least common denominator (LCD) for both expressions.
The first thing we want to do is to make sure that each denominator is factored completely. The second expression is, but the first is not.
We see that the denominator in the first expression is quadratic; therefore, it follows the formula:
$ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers.
To factor the expression in the denominator of the first fraction, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $-5$, but when added together will give us the $b$ term, which is $-4$. This means that one factor is positive and the other negative, with the negative factor having the greater absolute value.
Let's look at possible factors:
$-5$ and $1$
Let's rewrite the expression in factor form:
$(g - 5)(g + 1)$
The exercise can now be rewritten as:
$\frac{4g - 8}{(g - 5)(g + 1)} - \frac{3}{g - 5}$
Next, we want to find the least common denominator (LCD). We do this by taking the highest power of each factor in the denominators of the fractions. Before we can do that, we need to factor the denominators:
LCD = $(g - 5)(g + 1)$
Now that we have the least common denominator, we multiply the numerator of each fraction with the factor or factors it is missing in its denominator:
$\frac{4g - 8}{(g - 5)(g + 1)} - \frac{2(g + 1)}{(g - 5)(g + 1)}$
Rewrite the two fractions as one with the same denominator:
$\frac{(4g - 8) - [2(g + 1)]}{(g - 5)(g + 1)}$
Use the distributive property to rewrite the numerator:
$\frac{(4g - 8) - (2g + 2)}{(g - 5)(g + 1)}$
Combine like terms in the numerator:
$\frac{2g - 10}{(g - 5)(g + 1)}$
Factor out common terms in the numerator:
$\frac{2(g - 5)}{(g - 5)(g + 1)}$
Cancel out the $g - 5$ factor in the numerator and denominator:
$\frac{2}{g + 1}$
