- Last updated

- Save as PDF

- Page ID
- 66919

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vectorC}[1]{\textbf{#1}}\)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

##### Learning Objectives

- Use properties of equality together to isolate variables and solve algebraic equations.
- Use the properties of equality and the distributive property to solve equations containing parentheses, fractions, and/or decimals.

## Introduction

There are some **equations** that you can solve in your head quickly. For example, what is the value of \(\ y\) in the equation \(\ 2 y=6\)? Chances are you didn’t need to get out a pencil and paper to calculate that \(\ y=3\). You only needed to do one thing to get the answer, divide 6 by 2.

Other equations are more complicated. Solving \(\ 4\left(\frac{1}{3} t+\frac{1}{2}\right)=6\) without writing anything down is difficult! That’s because this equation contains not just a **variable** but also fractions and **terms** inside parentheses. This is a **multi-step equation**, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.

## Using Properties of Equalities

Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The **addition property of equality** and the **multiplication property of equality** explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal.

If the equation is in the form, \(\ a x+b=c\), where \(\ x\) is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.

##### Example

**Solve \(\ 3 y+2=11\).**

**Solution**

\(\ \begin{array}{r} 3 y+2=& 11 \\ \ -2 \ \ \ \ & \ -2 \\ \hline 3 y \ \ \ \ \ \ \ =&9 \end{array}\) | Subtract 2 from both sides of the equation to get the term with the variable by itself, so \(\ 3 y=9\). |

\(\ \frac{3 y}{3}=\frac{9}{3}\) \(\ y=3\) | Divide both sides of the equation by 3 to get a coefficient of 1 for the variable. |

\(\ y=3\)

##### Example

**Solve \(\ \frac{1}{4} x-2=3\).**

**Solution**

\(\ \begin{array}{r} \frac{1}{4} x-2=&3 \\ \ +2\ \ \ \ &\ +2 \\ \hline \frac{1}{4} x\ \ \ \ \ \ \ =&5 \end{array}\) | Add 2 to both sides of the equation to get the term with the variable by itself, so \(\ \frac{1}{4} x=5\). |

\(\ \begin{array}{c} \frac{4 x}{4}=5(4) \\ x=20 \end{array}\) | Multiply both sides of the equation by 4 to get a coefficient of 1 for the variable. |

\(\ x=20\)

If the equation is not in the form, \(\ a x+b=c\), you will need to perform some additional steps to get the equation in that form.

In the example below, there are several sets of **like terms**. You must first combine all like terms.

##### Example

**Solve \(\ 3 x+5 x+4-x+7=88\).**

**Solution**

\(\ 3 x+5 x+4-x+7=88\) | There are three like terms \(\ 3x\), \(\ 5x\) and \(\ -x\) involving a variable. |

\(\ \begin{array}{r} 7 x+4+7=88 \\ 7 x+11=88 \end{array}\) | Combine these like terms. 4 and 7 are also like terms and can be added. |

\(\ \begin{array}{r} 7 x+11= & 88 \\ -11\ \ \ \ & \ -11 \\ \hline \frac{7 x}{7}= & \frac{77}{7} \end{array}\) | The equation is now in the form \(\ a x+b=c\). So, we can solve \(\ 7 x+11=88\) as before. Subtract 11 from both sides. Divide both sides by 7. |

\(\ x=11\)

Some equations may have the variable on both sides of the equal sign. We need to “move” one of the variable terms in order to solve the equation.

##### Example

**Solve \(\ 6 x+5=10+5 x\). Check your solution.**

**Solution**

\(\ 6 x+5=10+5 x\) | This equation has \(\ x\) terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign. | |

\(\ \begin{array}{r} | You can subtract \(\ 5x\) on each side of the equal sign, which gives a new equation: \(\ x+5=10\). This is now a one-step equation! | |

\(\ \begin{array}{rr} x+5= & 10 \\ \ -5 \ \ \ \ & \ -5 \\ \hline x=&5 \end{array}\) | Subtract 5 from both sides and you get \(\ x=5\). | |

Check | \(\ \begin{aligned} 6 x+5 &=10+5 x \\ 6(5)+5 &=10+5(5) \\ 30+5 &=10+25 \\ 35 &=35 \end{aligned}\) | Check your solution by substituting 5 for \(\ x\) in the original equation. This is a true statement, so the solution is correct. |

\(\ x=5\)

Here are some steps to follow when you solve multi-step equations.

##### Solving multi-step equations

- If necessary, simplify the expressions on each side of the equation, including combining like terms.
- Get all variable terms on one side and all numbers on the other side using the addition property of equality. (\(\ a x+b=c \text { or } c=a x+b\))
- Isolate the variable term using the inverse operation or additive inverse (opposite) using the addition property of equality.
- Isolate the variable using the inverse operation or multiplicative inverse (reciprocal) using the multiplication property of equality to write the variable with a coefficient of 1.
- Check your solution by substituting the value of the variable in the original equation.

The examples below illustrate this sequence of steps.

##### Example

**Solve for \(\ y\).**

**\(\ -20 y+15=2-16 y+11\)**

**Solution**

\(\ \begin{array}{l} -20 y+15=2-16 y+11 \\ -20 y+15=-16 y+13 \end{array}\) | Step 1. On the right side, combine like terms: \(\ 2+11=13\). | |

\(\ \begin{array}{rr} -20 y+15= & -16 y+13 \\ +20 y \ \ \ \ \ \ \ \ \ \ \ \ \ & +20 y\ \ \ \ \ \ \ \ \ \\ \hline 15= & 4 y+13 \\ -13 \ \ \ \ & -13 \\ \hline 2= & 4 y\ \ \ \ \ \ \ \ \ \\ \frac{2}{4}= & \frac{4 y}{4} \ \ \ \ \ \ \ \ \\ \frac{1}{2}= & y\ \ \ \ \ \ \ \ \ \end{array}\) | Step 2. Add \(\ 20y\) to both sides to remove the variable term from the left side of the equation. Step 3. Subtract 13 from both sides. Step 4. Divide \(\ 4y\) by 4 to solve for \(\ y\). | |

Check | \(\ \begin{aligned} -20 y+15 &=2-16 y+11 \\ -20\left(\frac{1}{2}\right)+15 &=2-16\left(\frac{1}{2}\right)+11 \\ -10+15 &=2-8+11 \\ 5 &=5 \end{aligned}\) | Step 5. To check your answer, substitute \(\ \frac{1}{2}\) for \(\ y\) in the original equation. The statement \(\ 5=5\) is true, so \(\ y=\frac{1}{2}\) is the solution. |

\(\ y=\frac{1}{2}\)

##### Advanced Example

**Solve \(\ 3 y+10.5=6.5+2.5 y\). Check your solution.**

**Solution**

\(\ 3 y+10.5=6.5+2.5 y\) | This equation has \(\ y\) terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign. | |

\(\ \begin{array}{r} 3 y+10.5=&6.5+2.5 y \\ \ -2.5 y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\ -2.5 y \\ \hline 0.5 y+10.5=&6.5\ \ \ \ \ \ \ \ \ \ \ \ \end{array}\) | Add \(\ -2.5 y\) to both sides so that the variable remains on one side only. | |

\(\ \begin{array}{rrr} 0.5 y+10.5= & 6.5 \\ \ -10.5\ \ \ \ & \ -10.5 \\ \hline 0.5 y \ \ \ \ \ \ \ \ \ \ \ \ =&\ -4 \end{array}\) | Now isolate the variable by subtracting 10.5 from both sides. | |

\(\ \begin{aligned} 10(0.5 y) &=10(-4) \\ 5 y &=-40 \\ \frac{5 y}{5} &=\frac{-40}{5} \\ y &=-8 \end{aligned}\) | Multiply both sides by 10 so that \(\ 0.5y\) becomes \(\ 5y\), then divide by 5. | |

Check | \(\ \begin{aligned} 3 y+10.5 &=6.5+2.5 y \\ 3(-8)+10.5 &=6.5+(-20) \\ -24+10.5 &=6.5+(-20) \\ -13.5 &=-13.5 \end{aligned}\) | Check your solution by substituting -8 in for \(\ y\) in the original equation.his is a true statement, so the solution is correct. |

\(\ y=-8\)

##### Advanced Question

Identify the step that will *not* lead to a correct solution to the problem.

\(\ 3 a-\frac{11}{2}=-\frac{3 a}{2}+\frac{25}{2}\)

- Multiply both sides of the equation by 2.
- Add \(\ \frac{11}{2}\) to both sides of the equation.
- Add \(\ \frac{3 a}{2}\) to the left side, and add \(\ -3 a\) to the right side.
- Rewrite \(\ 3 a\) as \(\ \frac{6 a}{2}\).

**Answer**-
- Incorrect. Multiplying both sides by 2 keeps both sides equal; the new equation will be \(\ 6 a-11=-3 a+25\). This will lead to the correct solution. The step that will not lead to a correct solution is: Add \(\ \frac{3 a}{2}\) to the left side, and add \(\ -3 a\) to the right side.
- Incorrect. Adding \(\ \frac{11}{2}\) to both sides of the equation will keep both sides equal; the new equation will be \(\ 3 a=-\frac{3 a}{2}+\frac{36}{2}\). This will lead to the correct solution. The step that will not lead to a correct solution is: Add \(\ \frac{3 a}{2}\) to the left side, and add \(\ -3 a\) to the right side.
- Correct. Adding unequal amounts to the left and to the right will unbalance the equation, and you will no longer be able to solve accurately for \(\ a\).
- Incorrect. Rewriting \(\ 3 a\) as \(\ \frac{6 a}{2}\) does not change the value of the fraction at all, so this will keep both sides equal. The step that will not lead to a correct solution is: Add \(\ \frac{3 a}{2}\) to the left side, and add \(\ -3 a\) to the right side.

## Solving Equations Involving Parentheses, Fractions, and Decimals

More complex multi-step equations may involve additional symbols such as parentheses. The steps above can still be used. If there are parentheses, you use the distributive property of multiplication as part of Step 1 to simplify the expression. Then you solve as before.

##### The Distributive Property of Multiplication

For all real numbers \(\ a\), \(\ b\), and \(\ c\), \(\ a(b+c)=a b+a c\).

What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the routine steps described above to **isolate the variable** to solve the equation.

##### Example

**Solve for \(\ a\).**

**\(\ 4(2 a+3)=-3(a-1)+31\)**

**Solution**

\(\ \begin{aligned} 4(2 a+3) &=-3(a-1)+31 \\ 8 a+12 &=-3 a+3+31 \end{aligned}\) | Apply the distributive property to expand \(\ 4(2 a+3)\) to \(\ 8 a+12\) and \(\ -3(a-1)\) to \(\ -3 a+3\). |

\(\ \begin{array}{r} 8 a+12=&-3 a+34 \\ +3a\ \ \ \ \ \ \ \ \ \ \ \ \ &+3 a\ \ \ \ \ \ \ \ \ \\ \hline 11 a+12=&+34 \\ \ -12 \ \ \ \ &-12 \\ \hline \frac{11 a}{11}\ \ \ \ \ \ \ \ =&\frac{22}{11} \\ a=&2\ \ \ \ \ \ \ \ \ \ \ \end{array}\) | Combine like terms. Add \(\ 3a\) to both sides to move the variable terms to one side. Subtract 12 to isolate the variable term. Divide both terms by 11 to get a coefficient of 1. |

\(\ a=2\)

##### Example

In which of the following equations is the distributive property properly applied to the equation \(\ 2(y+3)=7\)?

- \(\ y+6=7\)
- \(\ 2 y+6=14\)
- \(\ 2 y+6=7\)
- \(\ 2 y+3=7\)

**Solution**

- Incorrect. All of the terms inside the parentheses must be multiplied by the value outside. The correct answer is \(\ 2 y+6=7\).
- Incorrect. When applying the distributive property, multiplication is spread only to the terms inside the parentheses, not to the other parts of the equation. The correct answer is \(\ 2 y+6=7\).
- Correct. Since the distributive property allows us to distribute the multiplication of an entire expression to each of the terms of the expression separately, \(\ 2 y+6=7\) is correct.
- Incorrect. All of the terms inside the parentheses must be multiplied by the value outside. The correct answer is \(\ 2 y+6=7\).

If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. See the example below.

##### Example

**Solve \(\ \frac{1}{2} x-3=2-\frac{3}{4} x\) by clearing the fractions in the equation first.**

**Solution**

\(\ \begin{aligned} \frac{1}{2} x-3 &=2-\frac{3}{4} x \\ 4\left(\frac{1}{2} x-3\right) &=4\left(2-\frac{3}{4} x\right) \\ 4\left(\frac{1}{2} x\right)-4(3) &=4(2)-4\left(\frac{3}{4} x\right) \\ \frac{4}{2} x-12 &=8-\frac{12}{4} x \end{aligned}\) | Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.se the distributive property to expand the expressions on both sides. Multiply. |

\(\ \begin{array}{r} 2 x-12= & 8-3 x \\ \ +3 x\ \ \ \ \ \ \ \ \ \ \ \ \ & \ +3 x \\ \hline 5 x-12=&8 \ \ \ \ \ \ \ \ \ \\ \ +12\ \ \ \ &\ +12\ \ \ \ \ \ \ \ \ \\ \hline \frac{5 x}{5} \ \ \ \ \ \ \ \ \ =&\frac{20}{5}\ \ \ \ \ \ \ \ \\ x=&4 \ \ \ \ \ \ \ \ \ \end{array}\) | Add \(\ 3x\) to both sides to move the variable terms to only one side. Add 12 to both sides to move the Divide to isolate the variable. |

\(\ x=4\)

Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.

##### Example

**Solve \(\ \frac{1}{2} x-3=2-\frac{3}{4} x\).**

**Solution**

\(\ \begin{array}{rr} \frac{1}{2} x-3= & 2-\frac{3}{4} x \\ \ +\frac{3}{4} x \ \ \ \ \ \ \ = & \ +\frac{3}{4} x \\ \hline \frac{5}{4} x-3= & 2 \ \ \ \ \ \ \ \ \ \ \\ +3\ \ \ \ & \ +3\ \ \ \ \ \ \ \ \ \ \\ \hline \frac{5}{4} x\ \ \ \ \ \ \ = & 5\ \ \ \ \ \ \ \ \ \ \end{array}\) | Add \(\ \frac{3}{4} x\) to both sides to get the variable terms on one side. \(\ \frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{5}{4}-\frac{3}{4}+\frac{3}{4}=0\) Add 3 to both sides to get the constant terms on the other side. |

\(\ \begin{aligned} \frac{4}{5} \cdot \frac{5}{4} x &=\frac{4}{5} \cdot 5 \\ x &=\frac{20}{5} \\ x &=4 \end{aligned}\) | To get a coefficient of 1, multiply the variable term by its multiplicative inverse. |

\(\ x=4\)

##### Advanced Example

**Solve \(\ \frac{1}{2}(2+a)=\frac{3 a+4}{3^{2}}\). Check your solution.**

**Solution**

\(\ \frac{1}{2}(2+a)=\frac{3 a+4}{3^{2}}\) | Solving this equation will require multiple steps. Begin by evaluating \(\ 3^{2}=9\). | |

\(\ \begin{aligned} \frac{1}{2} \cdot 2+\frac{1}{2} \cdot a &=\frac{3 a+4}{9} \\ 1+\frac{1}{2} a &=\frac{3 a+4}{9} \\ 1+\frac{a}{2} &=\frac{3 a+4}{9} \end{aligned}\) | Now distribute the \(\ \frac{1}{2}\) on the left side of the equation. | |

\(\ \begin{aligned} 18\left(1+\frac{a}{2}\right) &=18\left(\frac{3 a+4}{9}\right) \\ 18 \cdot 1+18 \cdot \frac{a}{2} &=\frac{18(3 a+4)}{9} \\ 18+\frac{18 a}{2} &=\frac{18(3 a+4)}{9} \\ 18+\frac{18 a}{2} &=\frac{9 \cdot 2(3 a+4)}{9} \\ 18+9 a \cdot \frac{2}{2} &=2(3 a+4) \cdot \frac{9}{9} \\ 18+9 a &=2(3 a+4) \\ 18+9 a &=2 \cdot 3 a+2 \cdot 4 \\ 18+9 a &=6 a+8 \end{aligned}\) | Multiply both sides of the equation by 18, the common denominator of the fractions in the problem. Use the distributive property to expand the expression on the left side. Then remove a factor of 1 from both sides. On the left, you can think of \(\ \frac{18 a}{2}\) as \(\ \frac{2}{2}\). On the right, you can think of \(\ \frac{18(3 a+4)}{9}\) as \(\ \frac{9}{9} \cdot \frac{2(3 a+4)}{1}\). Continue solving for \(\ a\) using the distributive property. | |

\(\ \begin{array}{rr} 18+9 a= & 6 a+8 \\ \ -6 a \ \ \ \ & \ -6 a\ \ \ \ \ \ \ \\ \hline 18+3 a= & 8 \\ \ -18\ \ \ \ \ \ \ \ \ \ \ \ \ & \ -18 \\ \hline 3 a= & -10 \\ \frac{3 a}{3}= & \frac{-10}{3} \\ a= & -\frac{10}{3} \end{array}\) | Then isolate the variable, and solve the remaining one-step problem. | |

Check | \(\ \begin{aligned} \frac{1}{2}\left[2+\left(-\frac{10}{3}\right)\right] &=\frac{3\left(-\frac{10}{3}\right)+4}{3^{2}} \\ \frac{1}{2}\left[2+\left(-\frac{10}{3}\right)\right] &=\frac{-10+4}{9} \\ \frac{1}{2}\left[\frac{6}{3}+\left(-\frac{10}{3}\right)\right] &=\frac{-10+4}{9} \\ \frac{1}{2}\left(-\frac{4}{3}\right) &=\frac{-10+4}{9} \\ \frac{-4}{6} &=\frac{-10+4}{9} \\ \frac{-4}{6} &=\frac{-6}{9} \\ -\frac{2}{3} \cdot \frac{2}{2} &=-\frac{2}{3} \cdot \frac{3}{3} \\ -\frac{2}{3} &=-\frac{2}{3} \end{aligned}\) | Check your solution by substituting \(\ -\frac{10}{3}\) in for \(\ a\) in the original equation. This is a true statement, so the solution is correct. |

\(\ a=-\frac{10}{3}\)

##### Exercise

To clear the fractions from \(\ \frac{1}{3}-\frac{2 y}{9}=19\), we can multiply both sides of the equation by which of the following numbers?

\(\ \begin{array}{llll}

3 & 6 & 9 & 27

\end{array}\)

- 9
- 9 and 27
- 6
- 3 or 9

**Answer**-
- Incorrect. While 9 is a common denominator of \(\ \frac{1}{3}\) and \(\ \frac{2}{9}\), so is 27. Any denominator will work, not just the least one. The correct answer is 9 and 27.
- Correct. Both 9 and 27 are common denominators of \(\ \frac{1}{3}\) and \(\ \frac{2}{9}\).
- Incorrect. You clear fractions by multiplying them by a common denominator. 6 is not a common denominator of \(\ \frac{1}{3}\) and \(\ \frac{2}{9}\). The correct answer is 9 and 27.
- Incorrect. While 9 is a common denominator of \(\ \frac{1}{3}\) and \(\ \frac{2}{9}\), 3 is not. The correct answer is 9 and 27.

Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find easier! Remember to check your answer by substituting your solution into the original equation.

Just as you can clear fractions from an equation, you can clear decimals from the equation in the same way. Find a common denominator and use the multiplication property of equality to multiply both sides of the equation.

##### Example

**Solve \(\ 0.4 x-0.25=1.75\) by clearing the decimals first.**

**Solution**

\(\ \begin{array}{r} 0.4 x-0.25=1.75 \\ 100(0.4 x-0.25)=100(1.75) \end{array}\) | \(\ 0.4\left(\frac{4}{10}\right)\) and \(\ 0.25\left(\frac{25}{100}\right)\) and \(\ 1.75\left(\frac{175}{100}\right)\) have a common denominator of 100. Multiply both sides by 100. | |

\(\ \begin{array}{r} 40 x-25 =&175 \\ \ +25 \ \ \ \ &\ +25 \\ \hline \frac{40 x}{40}\ \ \ \ \ \ \ \ \ =&\frac{200}{40} \\ x =&5\ \ \ \ \ \ \ \end{array}\) | Apply the distributive property to clear the parentheses. Solve as before. Add 25 to both sides. Divide both sides by 40. | |

Check: | \(\ \begin{array}{r} 0.4 x-0.25=1.75 \\ 0.4(5)-0.25=1.75 \\ 2-1.25=1.75 \\ 1.75=1.75 \end{array}\) | Substitute \(\ x=5\) into the original equation. Evaluate. The solution checks. |

\(\ x=5\)

##### Advanced Question

Solve for \(\ a\): \(\ \frac{1}{4}(a+3)=2-a\)

- \(\ a=2\)
- \(\ a=1\)
- \(\ a=0\)
- \(\ a=-2\)

**Answer**-
- Incorrect. Try multiplying both sides of the equation by 4, as \(\ 4 \cdot \frac{1}{4}=1\). The new equation will be \(\ 4\left[\frac{1}{4}(a+3)\right]=4(2-a)\), which reduces to \(\ a+3=8-4 a\). The correct answer is: \(\ a=1\).
- Correct. You can solve this equation by multiplying both sides by 4, since \(\ 4 \cdot \frac{1}{4}=1\). The resulting equation, \(\ a+3=4(2-a)\) can be rewritten as \(\ a+3=8-4 a\), and then \(\ 5 a=5\). You find that \(\ a=1\).
- Incorrect. Substituting \(\ a=0\) into the equation, you find \(\ \frac{1}{4}(0+3)=2-0\), so \(\ \frac{3}{4}=2\). This is not accurate. The correct answer is: \(\ a=1\).
- Incorrect. Try multiplying both sides of the equation by 4, as \(\ 4 \cdot \frac{1}{4}=1\). The new equation will be \(\ 4\left[\frac{1}{4}(a+3)\right]=4(2-a)\), which reduces to \(\ a+3=8-4 a\). The correct answer is: \(\ a=1\).

## Summary

Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses, or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques.