Chapter A.6, Problem 142AYU

To determine

To find: The equation $\frac{5}{x+3}+3=\frac{8+x}{x+3}$ has no solution, yet when we go through the process of solving it we obtain $x=-3$. Write a brief paragraph to explain what causes this to happen.

Answer to Problem 142AYU

$x=-3$ is a singular point. The point $x=-3$ is not defined in the equation, since when we substitute $x=-3$ the denominator becomes zero.

Explanation of Solution

$\frac{5}{x+3}+3=\frac{8+x}{x+3}$

$\frac{5\left(x+3\right)}{x+3}+3\left(x+3\right)=\frac{8+x\left(x+3\right)}{x+3}$

$5+3x+9=8+x$

$2x=8-14$

$2x=-6$

$x=-3$

$x=-3$ is a singular point. The point $x=-3$ is not defined in the equation, since when we substitute $x=-3$ the denominator becomes zero.