\displaystyle{x}={8} or \displaystyle{x}=-\frac{{2}}{{3}} (neither solution is extraneous) Explanation: Consider the two possibilities for \displaystyle{2}{x}-{3} : \displaystyle{\left.\begin{array}{ccc} \text{if }\ {2}{x}-{3}\ge{0}&{\left(\text{xxx}\right)}&\text{if }\ {2}{x}-{3}{<}{0}\\{\left|{{2}{x}-{3}}\right|}={x}+{5}&&{\left|{{2}{x}-{3}}\right|}={x}+{5}\\\rightarrow{2}{x}-{3}={x}+{5}&&\rightarrow{3}-{2}{x}={x}+{5}\\\rightarrow{x}={8}&&\rightarrow-{2}={3}{x}\\&&\rightarrow{x}=-\frac{{2}}{{3}}\end{array}\right.} ...

\displaystyle{x}=-{1} \displaystyle{x}={4} Explanation: As explained below: Step 1: Clear the absolute value bars \displaystyle{\left|{2}{x}-{3}\right|}={5} For negative part, lets ...

KillerBunny Feb 4, 2015 First of all, you have to determine the absolute value. Since \displaystyle{\left|{a}\right|}={a} if \displaystyle{a}{>}{0} and \displaystyle-{a} if \displaystyle{a}{<}{0} ...

\displaystyle{x}={5} and \displaystyle{x}={1} Explanation: First, isolate the absolute value term on the left side of the equation: \displaystyle{\left|{{x}-{3}}\right|}+{1}-{1}={3}-{1} ...

See a solution process below: Explanation: First, add \displaystyle{\left({6}\right)} to each side of the equation to isolate the absolute value function while keeping the equation balanced: \displaystyle{\left|{{x}-{3}}\right|}-{6}+{\left({6}\right)}={2}+{\left({6}\right)} ...

Don't Memorise Apr 29, 2015 We can divide both sides by 2 to get \displaystyle{\left|{{x}-{5}}\right|}{>}\frac{{32}}{{2}} \displaystyle\to{\left|{{x}-{5}}\right|}{>}{16} This ...