Chapter 7.1, Problem 35WE

To determine

Solve the equation $\frac{1}{y+2}+\frac{1}{y+6}=1$ and check for extraneous roots by substitution of answer into the original equation.

Answer to Problem 35WE

The solution set for the equation is $\left\{-3+\sqrt{5},-3-\sqrt{5}\right\}$ .

Explanation of Solution

Given information: The given equation is,

  $\frac{1}{y+2}+\frac{1}{y+6}=\mathrm{1....}\left(1\right)$

Calculation: Simplify (1) by multiplying both sides by $\left(y+2\right)\left(y+6\right)$ gives,

  $\begin{array}{c}\left(y+6\right)+\left(y+2\right)=\left(y+2\right)\left(y+6\right)\mathrm{....}\left(\begin{array}{l}\text{Multiply both sides}\\ \text{by }\left(y+2\right)\left(y+6\right)\end{array}\right)\\ 2y+8=y^2+8y+12\\ \text{or, }{y}^2+6y+4=0\end{array}$

Further solving,

  $\begin{array}{c}y=\frac{-\left(6\right)\pm \sqrt{{\left(6\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\mathrm{....}\left(\text{Quadratic formula}\right)\\ =\frac{-6\pm 2\sqrt{5}}{2}\mathrm{....}\left(\text{Simplified}\right)\\ =-3\pm \sqrt{5}\\ =-3+\sqrt{5},-3-\sqrt{5}\end{array}$

Since both the roots $-3+\sqrt{5},-3-\sqrt{5}$ satisfy (1) and do not make the denominator of (1) equal to zero.

Therefore, the solution set for the given equation is $\left\{-3+\sqrt{5},-3-\sqrt{5}\right\}$ .