Chapter 1.4, Problem 50E

To determine

To solve: the equation and correct the error.

Answer to Problem 50E

$x=-2$ and $x=-\frac{4}{3}$ is not a solution

Explanation of Solution

Given:

  $\begin{array}{l}\left|5x+8\right|=x\\ 5x+8=x\text{ or 5}x+8=-x\\ 4x+8=0\text{ 6}x+8=0\\ 4x=-8\text{ 6}x=-8\\ x=-2x=-\frac{4}{3}\\ \text{ The solution are }x=-2andx=-\frac{4}{3}\end{array}$

Calculation:

$\left|5x+8\right|=x$	given the equation:
$x=-2,x=-\frac{4}{3}$	Everything is correct except for the last line. In theprevious line, the possible solutions are obtatined:
$\begin{array}{l}x=-2\\ \left|5\left(-2\right)+8\right|=\left|-10+8\right|=\left|-2\right|=2\\ 2\cancel{=}-2\Rightarrow x=-2\text{ is not a solution}\\ x=-\frac{4}{3}\\ \left|5\left(-\frac{4}{3}\right)+8\right|=\left|-\frac{20}{3}+8\right|=\left|\frac{4}{3}\right|=\frac{4}{3}\\ \frac{4}{3}\cancel{=}-\frac{4}{3}\Rightarrow x=-\frac{4}{3}\text{ is not a solution}\end{array}$	check the solutions by substituting them inthe original equation:
	Observation: say from the very beginning, just looking at the equation, that $x$ must be positive because the left side of the equation is positive, so the domain of the equation is $[0,\infty )$ . As bothpossible solutions are negative, the equation has no solution even without checking the found values.

Conclusion:

  $x=-2$ and $x=-\frac{4}{3}$ is not a solution