This equation means that if you begin with some unknown number, x, and subtract 6, you will end up with 8. You are trying to figure out the value of the variable x.

Using the Addition Property of Equality, add 6 to both sides of the equation to isolate the variable. You choose to add 6 because 6 is being subtracted from the variable.

[latex]\displaystyle \begin{array}{r}x-6\,\,\,=\,\,\,\,8\\\,\,\,\,\,\,\,\underline{+\,6\,\,\,\,\,\,\,\,+6}\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,=\, 14\end{array}[/latex]

Answer

[latex]x=14[/latex]

This equation means that if you begin with some unknown number, x, and add 5, you will end up with 27. You are trying to figure out the value of the variable x.

Using the Addition Property of Equality, subtract 5 from both sides of the equation to isolate the variable. You choose to subtract 5, as 5 is being added from the variable.

[latex]\displaystyle \begin{array}{r}x+5\,\,=\,\,27\\\,\,\,\,\,\,\,\underline{-5\,\,\,\,\,\,\,\,-5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,=\, 22\end{array}[/latex]

Answer

[latex]x=22[/latex]

Solve:

[latex]x+10=-65[/latex]

Since 10 is being added to the variable, subtract 10 from both sides. Note that subtracting 10 is the same as adding [latex]–10[/latex].

[latex]\displaystyle \begin{array}{r}x+10\,\,=\,\,\,\,-65\\\,\,\,\,\,\underline{-10\,\,\,\,\,\,\,\,\,\,-10}\\\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,=\,\,\,-75\end{array}[/latex]

To check, substitute the solution, [latex]–75[/latex] for x in the original equation, then simplify.

[latex]\displaystyle \begin{array}{r}\,\,\,\,\,x+10\,\,\,=-65\\-75+\,10\,\,\,=-65\\\,\,\,\,\,\,\,\,\,\,\,-65\,\,\,=-65\end{array}[/latex]

This equation is true, so the solution is correct.

Answer

[latex]x=–75[/latex] is the solution to the equation [latex]x+10=–65[/latex].

Solve:

[latex]x-4=-32[/latex]

Since 4 is being subtracted from the variable, add 4 to both sides.

[latex]\displaystyle \begin{array}{r}x-4\,\,=\,\,\,\,-32\\\,\,\,\,\,\underline{+4\,\,\,\,\,\,\,\,\,\,+4}\\\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,=\,\,\,-28\end{array}[/latex]

Check:

To check, substitute the solution, [latex]–28[/latex] for x in the original equation, then simplify.

[latex]\displaystyle \begin{array}{r}\,\,\,\,\,x-4\,\,\,=-32\\-28-\,4\,\,\,=-32\\\,\,\,\,\,\,\,\,\,\,\,-32\,\,\,=-32\end{array}[/latex]

This equation is true, so the solution is correct.

Answer

[latex]x=–28[/latex] is the solution to the equation [latex]x-4=–32[/latex].

This equation asks you to find what number plus 5 has an absolute value of 15. Since 15 and [latex]−15[/latex] both have an absolute value of 15, the absolute value equation is true when the quantity [latex]x + 5[/latex] is 15 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 x will make this expression equal to 15 as well as what value for x 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]

Divide both sides of the equation by 3 to isolate the variable (have a coefficient of 1). Dividing by 3 is the same as having multiplied by [latex]\frac{1}{3}[/latex].

[latex]\begin{array}{r}\underline{3x}=\underline{24}\\3\,\,\,\,\,\,\,\,\,\,\,\,3\,\\x=8\,\,\,\end{array}[/latex]

Check by substituting your solution, 8, for the variable in the original equation.

[latex]\begin{array}{r}3x=24 \\ 3\cdot8=24 \\ 24=24\end{array}[/latex]

The solution is correct!

Answer

[latex]x=8[/latex]

The only difference between this and the previous example is that the coefficient on x is a fraction. In the last example we used the reciprocal of 3 to isolate the x (disguised as division). Now we can use the reciprocal of [latex]\frac{1}{2}[/latex], which is 2.

Multiply both sides by 2:

[latex]\begin{array}{r}\left(2\right)\frac{1}{2 }{ x }=\left(2\right){ 8}\\\left(2\right)\frac{1}{2 } = 1\,\,\,\,\,\,\,\,\\\left(2\right)8 = 16\,\,\,\,\,\\{ x }=16\,\,\,\,\,\end{array}[/latex]

We want to isolate the k, which is being divided by 10. The first thing we should do is multiply both sides by 10.

[latex]\begin{array}{l}\text{ multiply the left side by 10 }\\\\\left(10\right)-\frac{7}{2}=\frac{10\cdot7}{2} = \frac{70}{2} = 35\\\\\text{ now multiply the right side by 10 }\\\\\,\,\,\frac{k}{10}\left(10\right) = \frac{k\cdot10}{10} = k\\\\\text{ now replace your results in the equation }\\35=k\end{array}[/latex]

Answer

We write the k on the left side as a matter of convention.

[latex]k=35[/latex]

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

Since 6 and [latex]−6[/latex] both have an absolute value of 6, the absolute value equation is true when the quantity [latex]2x[/latex] is 6 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 x will make this expression equal to 6 as well as what value for x 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]x=36[/latex] and [latex]x=−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]

Subtract 2 from both sides of the equation to get the term with the variable by itself.

[latex]\displaystyle \begin{array}{r}3y+2\,\,\,=\,\,11\\\underline{\,\,\,\,\,\,\,-2\,\,\,\,\,\,\,\,-2}\\3y\,\,\,\,=\,\,\,\,\,9\end{array}[/latex]

Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.

[latex]\begin{array}{r}\,\,\,\,\,\,\underline{3y}\,\,\,\,=\,\,\,\,\,\underline{9}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\\\,\,\,\,\,\,\,\,\,\,y\,\,\,\,=\,\,\,\,3\end{array}[/latex]

Answer

[latex]y=3[/latex]

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

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

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

[latex]7x+11\,\,\,=\,\,\,88[/latex]

Subtract 11 from both sides.

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

Divide both sides by 7.

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

Answer

[latex]x=11[/latex]

Write the two equations that will give an absolute value of 26.

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

Solve each equation for p 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 5 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 3. 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 5 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]

Apply the distributive property to expand [latex]4\left(2a+3\right)[/latex] to [latex]8a+12[/latex]

[latex]\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}[/latex]

Subtract 12 from both sides to isolate the variable term.

[latex]\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}[/latex]

Divide both terms by 8 to get a coefficient of 1.

[latex]\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\\a\,=\,\,2\end{array}[/latex]

Answer

[latex]a=2[/latex]

Apply the distributive property to expand [latex]2\left(4t-5\right)[/latex] to [latex]8t-10[/latex] and [latex]-3\left(2t+1\right)[/latex] to[latex]-6t-3[/latex]. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.

[latex]\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Add [latex]-6t[/latex] to both sides to begin combining like terms.

[latex]\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}[/latex]

Add 10 to both sides of the equation to isolate t.

[latex]\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}[/latex]

The last step is to divide both sides by 14 to completely isolate t.

[latex]\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}[/latex]

Answer

[latex]t=\frac{1}{2}[/latex]

We simplified the fraction [latex]\frac{7}{14}[/latex] into [latex]\frac{1}{2}[/latex]

Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.

[latex]\begin{array}{r}\frac{1}{2}x-3=2-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\end{array}[/latex]

Use the distributive property to expand the expressions on both sides. Multiply.

[latex]\begin{array}{r}4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)-4\left(-\frac{3}{4}x\right)\\\\ \frac{4}{2}x-12=8-\frac{12}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}[/latex]

Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.

[latex]\begin{array}{r}2x-12=8-3x\, \\\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Add 12 to both sides to move the constant terms to the other side.

[latex]\begin{array}{r}5x-12=8\,\,\\ \underline{\,\,\,\,\,\,+12\,+12} \\5x=20\end{array}[/latex]

Divide to isolate the variable.

[latex]\begin{array}{r}\underline{5x}=\underline{5}\\ 5\,\,\,\,\,\,\,\,\,5\\ x=4\end{array}[/latex]

Answer

[latex]x=4[/latex]

Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0 and clear the decimals from the equation.

[latex]\begin{array}{r}3y+10.5=6.5+2.5y\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 10\left(3y+10.5\right)=10\left(6.5+2.5y\right)\end{array}[/latex]

Use the distributive property to expand the expressions on both sides.

[latex]\begin{array}{r}10\left(3y\right)+10\left(10.5\right)=10\left(6.5\right)+10\left(2.5y\right)\end{array}[/latex]

Multiply.

[latex]30y+105=65+25y[/latex]

Move the smaller variable term, [latex]25y[/latex], by subtracting it from both sides.

[latex]\begin{array}{r}30y+105=65+25y\,\,\\ \underline{-25y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25y} \\5y+105=65\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Subtract 105 from both sides to isolate the term with the variable.

[latex]\begin{array}{r}5y+105=65\,\,\,\\ \underline{\,\,\,\,\,\,-105\,-105} \\5y=-40\end{array}[/latex]

Divide both sides by 5 to isolate the y.

[latex]\begin{array}{l}\underline{5y}=\underline{-40}\\ 5\,\,\,\,\,\,\,\,\,\,\,\,\,5\\ \,\,\,x=-8\end{array}[/latex]

Answer

[latex]x=-8[/latex]