There are three like terms involving a variable: [latex]3x[/latex], [latex]5x[/latex], and [latex]–x[/latex]. Combine these like terms. [latex]4[/latex] and [latex]7[/latex] are also like terms and can be added.

[latex]\begin{array}{r}\,\,3x+5x+4-x+7=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x+11=\,\,\,88\end{array}[/latex]

The equation is now in the form [latex]ax+b=c[/latex], so we can solve as before.

Subtract [latex]11[/latex] from both sides.

[latex]\begin{array}{r}7x+11\,\,\,=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-11\,\,\,\,\,\,\,-11}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x\,\,\,=\,\,\,77\end{array}[/latex]

Divide both sides by [latex]7[/latex].

[latex]\begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{7x}\,\,\,=\,\,\,\underline{77}\\7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,=\,\,\,11\end{array}[/latex]

Choose the variable term to move—to avoid negative terms choose [latex]2x[/latex].

[latex]\begin{array}{r}4x-6=2x+10\,\\\underline{-2x\,\,\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,\,\,\,\,\,}\\2x-6=10\end{array}[/latex]

Now add [latex]6[/latex] to both sides to isolate the term with the variable.

[latex]\begin{array}{r}2x-6=10\\\underline{\,\,\,\,+6\,\,\,+6}\\2x=16\end{array}[/latex]

Now divide each side by [latex]2[/latex] to isolate the variable [latex]x[/latex].

[latex]\begin{array}{c}\Large\frac{2x}{2}\normalsize=\Large\frac{16}{2}\\\\\normalsize{x=8}\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!

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=\Large\frac{3}{2}\normalsize \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=\Large\frac{3}{2}\normalsize \,\,\,\,\,\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(\Large\frac{3}{2}\normalsize\right)-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\\,\,\,\,\,\,3\left|\Large\frac{12}{2}\normalsize -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!