This equation asks you to find what number plus [latex]5[/latex] has an absolute value of [latex]15[/latex]. Since 15 and [latex]−15[/latex] both have an absolute value of [latex]15[/latex], the absolute value equation is true when the quantity [latex]x + 5[/latex] is [latex]15[/latex] or [latex]x + 5[/latex] is [latex]−15[/latex], since [latex]|15|=15[/latex] and [latex]|−15|=15[/latex]. So, you need to find out what value for  [latex]x[/latex] will make this expression equal to [latex]15[/latex], as well as what value for [latex]x[/latex] will make the expression equal to [latex]−15[/latex]. Solving the two equations you get

[latex] \displaystyle \begin{array}{l}x+5=15\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,{x+5=-15}\\\underline{\,\,\,\,\,-5\,\,\,\,-5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-5\,\,\,\,\,\,\,\,\,-5}\\x\,\,\,\,\,\,\,\,\,=\,\,10\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,=-20\end{array}[/latex]

You can check these two solutions in the absolute value equation to see if [latex]x=10[/latex] and [latex]x = −20[/latex] are correct.

[latex] \displaystyle \begin{array}{r}\,\,\left| x+5 \right|=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| x+5 \right|=15\\\left| 10+5 \right|=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -20+5 \right|=15\\\,\,\,\,\,\,\,\left| 15 \right|=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -15 \right|=15\\\,\,\,\,\,\,\,\,\,15=15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15=15\end{array}[/latex]

This equation asks you to find what number times [latex]2[/latex] has an absolute value of [latex]6[/latex].

Since [latex]6[/latex] and [latex]−6[/latex] both have an absolute value of [latex]6[/latex], the absolute value equation is true when the quantity [latex]2x[/latex] is [latex]6[/latex] or [latex]2x[/latex] is [latex]−6[/latex], since [latex]|6|=6[/latex] and [latex]|−6|=6[/latex].

So, you need to find out what value for [latex]x[/latex] will make this expression equal to [latex]6[/latex], as well as what value for [latex]x[/latex] will make the expression equal to [latex]−6[/latex].

Solving the two equations you get

[latex]2x=6\text { or } 2x=-6[/latex]

[latex]\frac{2x}{2}=\frac{6}{2}\text { or } \frac{2x}{2}=\frac{-6}{2}[/latex]

[latex]x=3\text { or } x=-3[/latex]

You can check these two solutions in the absolute value equation to see if [latex]x=3[/latex] and [latex]x=−3[/latex] are correct.

[latex] \displaystyle \begin{array}{r}\,\,\left|3\cdot2 \right|=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -3\cdot2 \right|=6\\\left| 6 \right|=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -6 \right|=6\\\end{array}[/latex]

Notice how this example is different from the last; [latex] \displaystyle\frac{1}{3}[/latex] is outside the absolute value grouping symbols. This means we need to isolate the absolute value first, then apply the definition of absolute value.

First, isolate the absolute value term by multiplying by the inverse of [latex] \displaystyle\frac{1}{3}[/latex]:

[latex]\begin{array}{r}\frac{1}{3}\left|k\right|=12\,\,\,\,\,\,\,\,\\\left(3\right)\frac{1}{3}\left|k\right|=\left(3\right)12\\\left|k\right|=36\,\,\,\,\,\,\,\,\end{array}[/latex]

Apply the definition of absolute value:

[latex] \displaystyle{k }=36\text { or } {k }=-36[/latex]

You can check these two solutions in the absolute value equation to see if [latex]k=36[/latex] and [latex]k=−36[/latex] are correct.

[latex] \displaystyle \begin{array}{r}\,\,\frac{1}{3}\left|36 \right|=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}\left|-36 \right|=12\\\left| 12 \right|=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -12 \right|=12\\\end{array}[/latex]

Write the two equations that will give an absolute value of [latex]26[/latex].

[latex] \displaystyle 2p-4=26\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,2p-4=\,-26[/latex]

Solve each equation for [latex]p[/latex] by isolating the variable.

[latex] \displaystyle \begin{array}{r}2p-4=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2p-4=\,-26\\\underline{\,\,\,\,\,\,+4\,\,\,\,+4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,+4\,\,\,\,\,\,\,+4}\\\underline{2p}\,\,\,\,\,\,=\underline{30}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{2p}\,\,\,\,\,=\,\underline{-22}\\2\,\,\,\,\,\,\,=\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,2\\\,\,\,\,\,\,\,\,\,p=15\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=\,-11\end{array}[/latex]

Check the solutions in the original equation.

[latex] \displaystyle \begin{array}{r}\,\,\,\,\,\left| 2p-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 2p-4 \right|=26\\\left| 2(15)-4 \right|=26\,\,\,\,\,\,\,\left| 2(-11)-4 \right|=26\\\,\,\,\,\,\left| 30-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -22-4 \right|=26\\\,\,\,\,\,\,\,\,\,\,\,\,\left| 26 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -26 \right|=26\end{array}[/latex]

Both solutions check!

Answer

[latex]p=15[/latex] or [latex]p=-11[/latex]

Isolate the term with the absolute value by adding  [latex]5[/latex] to both sides.

[latex]\begin{array}{r}3\left|4w-1\right|-5=10\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,+5\,\,\,+5}\\ 3\left|4w-1\right|=15\end{array}[/latex]

Divide both sides by [latex]3[/latex]. Now the absolute value is isolated.

[latex]\begin{array}{r} \underline{3\left|4w-1\right|}=\underline{15}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\\\left|4w-1\right|=\,\,5\end{array}[/latex]

Write the two equations that will give an absolute value of  [latex]5[/latex] and solve them.

[latex] \displaystyle \begin{array}{r}4w-1=5\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,4w-1=-5\\\underline{\,\,\,\,\,\,\,+1\,\,+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,\,\,+1\,\,\,\,\,+1}\\\,\,\,\,\,\underline{4w}=\underline{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{4w}\,\,\,\,\,\,\,=\underline{-4}\\4\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\\\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\text{or}\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Check the solutions in the original equation.

[latex] \displaystyle \begin{array}{r}\,\,\,\,\,3\left| 4w-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\3\left| 4\left( \frac{3}{2} \right)-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\\,\,\,\,\,\,3\left| \frac{12}{2}-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4(-1)-1\, \right|-5=10\\\\\,\,\,\,\,\,\,\,3\left| 6-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -4-1\, \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left(5\right)-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -5 \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\\10=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10=10\end{array}[/latex]

Both solutions check

Answer

[latex]w=-1\,\,\,\,\text{or}\,\,\,\,w=\frac{3}{2}[/latex]