Chapter 1.4, Problem 46HP

To determine

To solve: the given equation.

Answer to Problem 46HP

The result is $x\in \left\{-1,1\right\}$ .

Explanation of Solution

Given information:

Given equation is

  $\left|2x-1\right|+3=\left|5-x\right|$

Calculation:

In order to clear the absolute value bar, write all the possibilities:

  $\begin{array}{c}\left|2x-1\right|=\{\begin{array}{l}-2x+1,2x-1<0\\ 2x-1,2x-1\ge 0\end{array}\\ \left|5-x\right|=\{\begin{array}{l}-5+x,5-x<0\\ 5-x,5-x\ge 0\end{array}\end{array}$

Therefore,

  $\begin{array}{c}\left|2x-1\right|=\{\begin{array}{l}-2x+1,x<\frac{1}{2}\\ 2x-1,x\ge \frac{1}{2}\end{array}\\ \left|5-x\right|=\{\begin{array}{l}-5+x,x>5\\ 5-x,x\le 5\end{array}\end{array}$

The following key points and interval

  $\begin{array}{l}\left(\frac{1}{2},5\right)\\ \left(-\infty ,\frac{1}{2}\right),\left[\frac{1}{2},5\right],\left(5,\infty \right)\end{array}$

Solve the equation on given interval $x\in \left(-\infty ,\frac{1}{2}\right)$ ,

Now solve the equation,

  $\begin{array}{c}-2x+1+3=5-x\\ -2x+4+2x=5-x+2x\\ 4=5+x\\ x=-1\end{array}$

Now, check the solution is in the interval $\left(-\infty ,\frac{1}{2}\right)$ .

Solve the equation on given interval $x\in \left[\frac{1}{2},5\right]$ .

Now, solve the equation

  $\begin{array}{c}2x-1+3=5-x\\ 2x+2+x=5-x+x\\ 3x=3\\ x=1\end{array}$

Now, check the solution is in the interval $\left[\frac{1}{2},5\right]$

  $1\in \left[\frac{1}{2},5\right]$

Solve the equation of the given interval $x\in \left(5,\infty \right)$

Now, solve the equation

  $\begin{array}{c}2x-1+3=-5+x\\ 2x+2-x=5+x-x\\ x+2=-5\\ x=-7\end{array}$

Now, check the solution is in the interval $\left(5,\infty \right)$ .

  $-7\in \left(5,\infty \right)$

Hence, the result is $x\in \left\{-1,1\right\}$ .