Question

In the following exercise, find

(a) $R\left(x\right)=f\left(x\right)+g\left(x\right)$

(b) $R\left(x\right)=f\left(x\right)-g\left(x\right)$:

$$\begin{gathered} f\left(x\right)=\frac{-5x-5}{x^2+x-6} \\ g\left(x\right)=\frac{x+1}{2-x} \end{gathered}$$

Short answer

(a) $R\left(x\right)=-\frac{\left(x+8\right)\left(x+1\right)}{\left(x+3\right)\left(x-2\right)}$

(b)$R\left(x\right)=\frac{x+1}{x+3}$

Step-by-step solution

Step 1. Given information.

The following information is given:

$$\begin{gathered} f\left(x\right)=\frac{-5x-5}{x^2+x-6} \\ g\left(x\right)=\frac{x+1}{2-x} \end{gathered}$$

Step 2. Finding (a) Rx=fx+gx.

$R\left(x\right)=f\left(x\right)+g\left(x\right)$

Substituting the values of $f\left(x\right)$and $g\left(x\right)$:

$R\left(x\right)=\frac{-5x-5}{x^2+x-6}+\frac{x+1}{2-x}$

Factorise the denominators:

$R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}+\frac{\left(x+1\right)}{\left(2-x\right)}$

Multiply the numerator and denominator of the second term by $-1$:

$$\begin{gathered} R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}+\frac{\left(-1\right).\left(x+1\right)}{\left(-1\right).\left(2-x\right)} \\ R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}-\frac{x+1}{\left(x-2\right)} \end{gathered}$$

Find the LCD:

$\left(x+3\right)\left(x-2\right)$

Rewrite the expression with the numerators equal to the LCD:

$R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}$

Simplify:

$$\begin{gathered} R\left(x\right)=\frac{-5x-5-\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)} \\ R\left(x\right)=\frac{-5x-5-x^2-4x-3}{\left(x+3\right)\left(x-2\right)} \\ R\left(x\right)=-\frac{x^2+9x+8}{\left(x+3\right)\left(x-2\right)} \end{gathered}$$

Factorise the numerator:

$R\left(x\right)=-\frac{\left(x+8\right)\left(x+1\right)}{\left(x+3\right)\left(x-2\right)}$

Step 3. Finding (b) R(x)=fx-gx.

Substituting the values of $f\left(x\right)$and $g\left(x\right)$:

$R\left(x\right)=\frac{-5x-5}{x^2+x-6}-\frac{x+1}{2-x}$

Factorise the numerators:

$R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}-\frac{x+1}{2-x}$

Multiply the numerator and the denominator of both the terms with $-1$:

$R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}-\frac{\left(-1\right)\left(x+1\right)}{\left(-1\right)\left(2-x\right)}$

Simplify:

$R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}+\frac{\left(x+1\right)}{\left(x-2\right)}$

Find out the LCD:

$\left(x+3\right)\left(x-2\right)$

Rewrite the expressions:

role="math" localid="1644740490947" $R\left(x\right)=\frac{-5x-5}{\left(x+3\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}$

Simplify:

$$\begin{gathered} R\left(x\right)=\frac{-5x-5+x^2+4x+3}{\left(x+3\right)\left(x-2\right)} \\ R\left(x\right)=\frac{x^2-x-2}{\left(x+3\right)\left(x-2\right)} \end{gathered}$$

Factorise the numerator, cancel out common terms and simplify:

$$\begin{gathered} R\left(x\right)=\frac{\left(x-2\right)\left(x+1\right)}{\left(x+3\right)\left(x-2\right)} \\ R\left(x\right)=\frac{x+1}{x+3} \end{gathered}$$