Chapter 11.5, Problem 17P

Solve each equation. Check your solutions. If there is no solution, write no solution.

$\frac{2-2p}{p^2-6p+8}+\frac{3p}{p-4}=\frac{p}{p-2}$

To determine

To solve the equation and check the solution.

Answer to Problem 17P

The solution is $p=1$ .

Explanation of Solution

Given information:

The given equation is:

  $\frac{2-2p}{p^2-6p+8}+\frac{3p}{p-4}=\frac{p}{p-2}$

Calculation:

The solution can be obtained by:

  $\begin{array}{l}\frac{2-2p}{p^2-6p+8}+\frac{3p}{p-4}=\frac{p}{p-2}\\ \frac{2-2p}{p^2-4p-2p+8}+\frac{3p}{p-4}=\frac{p}{p-2}\\ \frac{2-2p}{p\left(p-4\right)-2\left(p-4\right)}+\frac{3p}{p-4}=\frac{p}{p-2}\\ \frac{2-2p}{\left(p-4\right)\left(p-2\right)}+\frac{3p}{p-4}=\frac{p}{p-2}\\ \frac{2-2p+3p\left(p-2\right)}{\left(p-4\right)\left(p-2\right)}=\frac{p}{p-2}\\ \frac{2-2p+3p^2-6p}{\left(p-4\right)\left(p-2\right)}=\frac{p}{p-2}\\ \frac{\left(2-8p+3p^2\right)}{\left(p-4\right)\left(p-2\right)}=\frac{p}{p-2}\\ \frac{\left(2-8p+3p^2\right)\left(p-2\right)}{\left(p-4\right)\left(p-2\right)}=p\\ 2-8p+3p^2=p\left(p-4\right)\\ 2-8p+3p^2=p^2-4p\\ 2-8p+3p^2-p^2+4p=0\\ 2p^2-4p+2=0\\ p^2-2p+1=0\\ p^2-p-p+1=0\\ p\left(p-1\right)-1\left(p-1\right)=0\\ \left(p-1\right)\left(p-1\right)=0\\ p=1\end{array}$

Hence, the solution is $p=1$ .

Put the value of $p$ in the solution to check the solution.