Chapter 11.5, Problem 37PPE

To determine

To solve the equation $\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$ and check the solution.

Answer to Problem 37PPE

  $s=-1,-\frac{6}{5}$

Explanation of Solution

Given:

The given equation is $\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$

Concept used :

• Least Common Denominator (LCD) is the smallest number that is divisible by the denominator of all the fractions that are being added or subtracted.
• Solving an equation containing rational expressions
1. Multiply both sides of equations by LCD(least common denominator) of all denominators.
2. Remove any grouping symbols and solve the resulting equation for the variable asked in equation.

Calculations :

In order to solve the equation $\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$ , observe that the LCD of all the denominators is $\left(3s+2\right)\cdot \left(2s-4\right)$ . So multiply both sides by this LCD to get rid of the denominators and simplify further as shown below

  $\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$

  $\frac{s}{3s+2}\cdot \left(3s+2\right)\cdot \left(2s-4\right)+\frac{s+3}{2s-4}\cdot \left(3s+2\right)\cdot \left(2s-4\right)=\frac{-2s}{3s^2-4s-4}\cdot \left(3s+2\right)\cdot \left(2s-4\right)$

  $s\cdot \left(2s-4\right)+\left(s+3\right)\cdot \left(3s+2\right)=\frac{-2s\cdot \left(3s+2\right)\cdot \left(2s-4\right)}{3s^2-\left(6-2\right)s-4}$

  $s\cdot \left(2s-4\right)+\left(s+3\right)\cdot \left(3s+2\right)=\frac{-2s\cdot \left(3s+2\right)\cdot \left(2s-4\right)}{3s^2-6s+2s-4}$

  $2s^2-4s+3s^2+2s+9s+6=\frac{-2s\cdot \left(3s+2\right)\cdot \left(2s-4\right)}{3s\left(s-2\right)+2\left(s-2\right)}$

  $5s^2+7s+6=\frac{-2s\cdot \left(3s+2\right)\cdot 2\left(s-2\right)}{\left(3s+2\right)\cdot \left(s-2\right)}$

  $5s^2+7s+6=-4s$

  $\begin{array}{c}5s^2+7s+4s+6=0\\ 5s^2+11s+6=0\\ 5s^2+\left(5+6\right)s+6=0\end{array}$

  $\begin{array}{c}5s^2+5s+6s+6=0\\ 5s\left(s+1\right)+6\left(s+1\right)=0\\ \left(5s+6\right)\cdot \left(s+1\right)=0\end{array}$

  $\begin{array}{c}5s+6=0\\ 5s=-6\\ s=-\frac{6}{5}\end{array}$

  $\begin{array}{c}s+1=0\\ s=-1\\ s=-1,-\frac{6}{5}\end{array}$

Thus, the solution of the given equation is $s=-1,-\frac{6}{5}$ .

Now to check the solution, substitute $s=-1,-\frac{6}{5}$ in the given equation and check whether left hand side value is equal to the right hand side value or no. So, it gives

  $\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$

  $\frac{-1}{3\cdot \left(-1\right)+2}+\frac{-1+3}{2\cdot \left(-1\right)-4}=\frac{-2\cdot \left(-1\right)}{3\cdot {\left(-1\right)}^2-4\cdot \left(-1\right)-4}$

  $\frac{-1}{-3+2}+\frac{2}{-2-4}=\frac{2}{3+4-4}$

  $\frac{-1}{-1}+\frac{2}{-6}=\frac{2}{3}$

  $1-\frac{1}{3}=\frac{2}{3}$

  $1\cdot 3-\frac{1}{3}\cdot 3=\frac{2}{3}\cdot 3$

  $\begin{array}{c}3-1=2\\ 2=2\\ ifs=-\frac{6}{5}\end{array}$

  $\frac{-\frac{6}{5}}{3\left(-\frac{6}{5}\right)+2}+\frac{-\frac{6}{5}+3}{2\left(-\frac{6}{5}\right)-4}=\frac{-2\left(-\frac{6}{5}\right)}{3{\left(-\frac{6}{5}\right)}^2-4\left(-\frac{6}{5}\right)-4}$

  $\frac{-\frac{6}{5}}{-\frac{18}{5}+2}+\frac{-\frac{6}{5}+3}{-\frac{12}{5}-4}=\frac{\frac{12}{5}}{\frac{108}{25}+\frac{24}{5}-4}$

  $\frac{-\frac{6}{5}}{-\frac{18}{5}+2\cdot \frac{5}{5}}+\frac{-\frac{6}{5}+3\cdot \frac{5}{5}}{-\frac{12}{5}-4\cdot \frac{5}{5}}=\frac{\frac{12}{5}}{\frac{108}{25}+\frac{24}{5}-4\cdot \frac{5}{5}}$

  $\frac{-\frac{6}{5}}{\frac{-18+10}{5}}+\frac{\frac{-6+15}{5}}{\frac{-12-20}{5}}=\frac{\frac{12}{5}}{\frac{108}{25}+\frac{24-20}{5}}$

  $\frac{6}{8}-\frac{9}{32}=\frac{\frac{12}{5}\cdot \frac{5}{5}}{\frac{108}{25}+\frac{4}{5}\cdot \frac{5}{5}}$

  $\begin{array}{c}\frac{6}{8}-\frac{9}{32}=\frac{\frac{60}{25}}{\frac{108}{25}+\frac{20}{25}}\\ \frac{6}{8}-\frac{9}{32}=\frac{60}{128}\\ \frac{6}{8}\cdot 32-\frac{9}{32}\cdot 32=\frac{15}{32}\cdot 32\\ 24-9=15\\ 15=15\text{True Statement}\end{array}$