Chapter 6.6, Problem 29WE

To determine

Simplify: the given expression as fraction.

Answer to Problem 29WE

The simplified form of $3x-\frac{x^2}{x+3}-2$ is $\frac{2x^2+7x-6}{x+3}$.

Explanation of Solution

Given information:

The expression $3x-\frac{x^2}{x+3}-2$.

Formula used:

Least common denominator: It is the smallest multiple of denominator of set of fractions.

  $\left(a+b\right)\left(a-b\right)=a^2-b^2$

  $3x-\frac{x^2}{x+3}-2$

Recall that Least common denominator is the smallest multiple of denominator of set of fractions.

  $=\frac{3x}{1}-\frac{x^2}{x+3}-\frac{2}{1}$

Least common denominator for $\frac{3x}{1},\frac{x^2}{x+3},\frac{2}{1}$ is $x+3$ . So,

  $\begin{array}{l}=\frac{3x\left(x+3\right)-x^2-2\left(x+3\right)}{x+3}\\ =\frac{3x^2+9x-x^2-2x-6}{x+3}\\ =\boxed{\frac{2x^2+7x-6}{x+3}}\end{array}$

The simplified form of $3x-\frac{x^2}{x+3}-2$ is $\frac{2x^2+7x-6}{x+3}$.