Chapter 12.3, Problem 28WE

To determine

To solve: the equation $\frac{w+3}{w-1}-\frac{7}{w+5}=3$

Answer to Problem 28WE

The required answer is: solution set is -7.86, 2.36

Explanation of Solution

Given information:

The given equation: $\frac{w+3}{w-1}-\frac{7}{w+5}=3$

Calculation:

  $\begin{array}{l}\frac{w+3}{w-1}-\frac{7}{w+5}=3\\ \frac{w+3}{w-1}-\frac{7}{w+5}+\frac{7}{w+5}=3+\frac{7}{w+5}\end{array}$

(Add $\frac{7}{w+5}$ to both sides)

  $\begin{array}{l}\frac{w+3}{w-1}=3+\frac{7}{w+5}\\ (\frac{w+3}{w-1})\cdot w-1=(3+\frac{7}{w+5})\cdot w-1\end{array}$

(Multiply $w-1$ on both sides)

  $\begin{array}{l}w+3=3w-3+\frac{7w-7}{w+5}\\ w+3-3w=3w-3+\frac{7w-7}{w+5}-3w\end{array}$

(Subtract 3w from both sides)

  $\begin{array}{l}3-2w=-3+\frac{7w-7}{w+5}\\ 3-2w+3=-3+\frac{7w-7}{w+5}+3\end{array}$

(Add 3 to both sides)

  $\begin{array}{l}-2w+6=\frac{7w-7}{w+5}\\ (-2w+6)\cdot w+5=(\frac{7w-7}{w+5})\cdot w+5\end{array}$

(Multiply $w+5$ to both sides)

  $\begin{array}{l}-2w^2+6w-10w+30=7w-7\\ -2w^2-4w+30-7w=7w-7-7w\end{array}$

(Subtract 7w from both sides)

  $\begin{array}{l}-2w^2+30-11w=-7\\ -2w^2+30-11w+7=-7+7\end{array}$

(Add 7 to both sides)

  $-2w^2-11w+37=0$

(Simplify)

  $w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=-2,b=-11andc=37$

  $w=\frac{-(-11)\pm \sqrt{{(-11)}^2-4(-2)(37)}}{2(-2)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{11\pm \sqrt{121+296}}{-4}\\ =\frac{11\pm \sqrt{417}}{-4}\end{array}$

(Simplify)

Since $\sqrt{417}\approx 20.42$ , $w\approx \frac{11+20.42}{-4}=\frac{31.42}{-4}\approx -7.86$

Or

  $w\approx \frac{11-20.42}{-4}=\frac{-9.42}{-4}\approx 2.36$

Therefore the solution set is $\frac{11+\sqrt{417}}{-4},\frac{11-\sqrt{417}}{-4}$ or - 7.86, 2.36