Subtract the following rational expressions.(1 point)

Question

x+1x+3−52x+6=
$$

Answer 1

To subtract the rational expressions, we need to find a common denominator.

The first rational expression is $\frac{x+1}{x+3}$.

The second rational expression is $\frac{5}{2x+6}$.

To find a common denominator, we need to factor the denominators.

The first denominator, $x+3$, is already factored.

The second denominator, $2x+6$, can be factored out the common factor of 2, giving $2(x+3)$.

So the common denominator is $(x+3)(2(x+3))$.

We can now rewrite the rational expressions with the common denominator.

The first rational expression becomes $\frac{(x+1)(2(x+3))}{(x+3)(2(x+3))}$.

The second rational expression remains $\frac{5}{2(x+3)}$.

Now we can subtract the rational expressions:

$\frac{(x+1)(2(x+3))}{(x+3)(2(x+3))} - \frac{5}{2(x+3)}$

To subtract rational expressions, we need a common denominator. In this case, the common denominator is $(x+3)(2(x+3))$.

Multiplying the first rational expression by $\frac{2(x+3)}{2(x+3)}$ and the second rational expression by $\frac{(x+3)(2(x+3))}{(x+3)(2(x+3))}$, we have:

$\frac{(x+1)(2(x+3)) \cdot 2(x+3)}{(x+3)(2(x+3)) \cdot 2(x+3)} - \frac{5 \cdot (x+3)(2(x+3))}{2(x+3) \cdot (x+3)(2(x+3))}$

Simplifying, we get:

$\frac{2(x+1)(x+3)(x+3)}{2(x+3)(x+3)(2(x+3))} - \frac{10(x+3)(x+3)}{2(x+3)(x+3)(2(x+3))}$

Now, both rational expressions have the same denominator. We can subtract them:

$\frac{2(x+1)(x+3)(x+3) - 10(x+3)(x+3)}{2(x+3)(x+3)(2(x+3))}$

Simplifying, we have:

$\frac{(2x+2)(x+3)(x+3) - 10(x+3)(x+3)}{2(x+3)(x+3)(2(x+3))}$

Expanding and simplifying further:

$\frac{(2x^2 + 8x + 6)(x+3) - (10x^2 + 60x + 90)}{2(x+3)(x+3)(2(x+3))}$

Distributing and combining like terms:

$\frac{2x^3 + 14x^2 + 26x - 84}{2(x+3)(x+3)(2(x+3))}$

Therefore, the simplified expression is $\frac{2x^3 + 14x^2 + 26x - 84}{2(x+3)(x+3)(2(x+3))}$