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3.3: Multi-Step Equations

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Solve a multi-step equation by combining like terms.
  • Solve a multi-step equation using the distributive property.
  • Solve real-world problems using multi-step equations.

Solving Multi-Step Equations by Combining Like Terms

We have seen that when we solve for an unknown variable, it can be a simple matter of moving terms around in one or two steps. We can now look at solving equations that take several steps to isolate the unknown variable. Such equations are referred to as multi-step equations.

In this section, we will simply be combining the steps we already know how to do. Our goal is to end up with all the constants on one side of the equation and all of the variables on the other side. We will do this by collecting “like terms”. Don’t forget, like terms have the same combination of variables in them.

Example 1

Solve \begin{align*} \frac{3x+4} {3} - 5x = 6\end{align*}3x+435x=6

This problem involves a fraction. Before we can combine the variable terms we need to deal with it. Let’s put all the terms on the left over a common denominator of three.

\begin{align*}& \frac{3x+4} {3} - \frac{15x} {3} =6\ && \text{Next we combine the fractions}. \\ & \frac{3x+4 - 15x} {3} = 6\ && \text{Combine like terms}. \\ & \qquad \ \frac{4 - 12x} {3} = 6\ && \text{Multiply both sides by}\ 3. \\ & \qquad \ \ \cancel{4}-12x =18\ && \text{Subtract}\ 4\ \text{from both sides}. \\ & \quad \ \underline{\;\; -\cancel{4} \qquad \quad \ \ -4 }\\ & \qquad \quad -12x =14\ && \text{Divide both sides by}\ -12 \\ & \qquad \quad \ \frac{-12} {-12} x = - \frac{14} {12}\end{align*}3x+4315x3=6 3x+415x3=6  412x3=6   412x=18  4  412x=14  1212x=1412Next we combine the fractions.Combine like terms.Multiply both sides by 3.Subtract 4 from both sides.Divide both sides by 12


\begin{align*} x = - \frac{7} {6}\end{align*}x=76

Solving Multi-Step Equations Using the Distributive Property

You have seen in some of the examples that we can choose to divide out a constant or distribute it. The choice comes down to whether on not we would get a fraction as a result. We are trying to simplify the expression. If we can divide out large numbers without getting a fraction, then we avoid large coefficients. Most of the time, however, we will have to distribute and then collect like terms.

Example 2

Solve \begin{align*}17(3x+ 4) =7\end{align*}17(3x+4)=7

This equation has the \begin{align*}x\end{align*}x buried in parentheses. In order to extract it we can proceed in one of two ways. We can either distribute the seventeen on the left, or divide both sides by seventeen to remove it from the left. If we divide by seventeen, however, we will end up with a fraction. We wish to avoid fractions if possible!

\begin{align*}& 17(3x+ 4) =7\ && \text{Distribute the}\ 17. \\ & \quad \ 51x \cancel{+68} = 7 \\ & \qquad \quad \cancel{-68} \ \ -68\ && \text{Subtract}\ 68\ \text{from both sides}. \\ & \qquad \quad 51x =-61\ && \text{Divide by}\ 51.\end{align*}17(3x+4)=7  51x+68=768  68 51x=61 Distribute the 17.Subtract 68 from both sides.Divide by 51.


\begin{align*} x = -\frac{61} {51}\end{align*}x=6151

Example 3

Solve \begin{align*}4(3x-4) - 7(2x+3) = 3\end{align*}4(3x4)7(2x+3)=3

This time we will need to collect like terms, but they are hidden inside the brackets. We start by expanding the parentheses.

\begin{align*}12x - 16 - 14x - 21 & = 3\ && \text{Collect the like terms}\ (12x\ \text{and}\ -14x). \\ (12x -14x) + (-16 - 21) & = 3\ && \text{Evaluate each set of like terms}. \\ -2x\cancel{-37} & =3 \\ \cancel{+37} & \ \ +37 && \text{Add} \ 37\ \text{to both sides}. \\ -2x& =40 \\ \frac{-2x}{-2} & =\frac{40}{-2}\ && \text{Divide both sides by}\ -2.\end{align*}12x1614x21(12x14x)+(1621)2x37+372x2x2=3 =3 =3  +37=40=402 Collect the like terms (12x and 14x).Evaluate each set of like terms.Add 37 to both sides.Divide both sides by 2.



Example 4

Solve the following equation for \begin{align*}x\end{align*}.

\begin{align*} 0.1(3.2 + 2x) + \frac{1} {2} \left (3- \frac{x} {5}\right ) = 0\end{align*}

This function contains both fractions and decimals. We should convert all terms to one or the other. It is often easier to convert decimals to fractions, but the fractions in this equation are easily moved to decimal form. Decimals do not require a common denominator!

Rewrite in decimal form.

\begin{align*}0.1 (3.2 +2x) +0.5 (3- 0.2x)& = 0\ && \text{Multiply out decimals:} \\ 0.32 +0.2x +1.5 - 0.1x & = 0\ && \text{Collect like terms:} \\ (0.2x -0.1x) + (0.32+1.5) & = 0\ && \text{Evaluate each collection:} \\ 0.1x \cancel{+1.82} & = 0\ && \text{Subtract}\ 1.82\ \text{from both sides} \\ \cancel{-1.82} & \ \ -1.82 \\ 0.1x & = -1.82\ && \text{Divide by}\ -0.1 \\ \frac{0.1x}{0.1} & = \frac{-1.82}{0.1}\end{align*}



Solve Real-World Problems Using Multi-Step Equations

Real-world problems require you to translate from a problem in words to an equation. First, look to see what the equation is asking. What is the unknown you have to solve for? That will determine the quantity we will use for our variable. The text explains what is happening. Break it down into small, manageable chunks, and follow what is going on with our variable all the way through the problem.

Example 5

A grower’s cooperative has a farmer’s market in the town center every Saturday. They sell what they have grown and split the money into several categories. 8.5% of all the money taken is removed for sales tax. $150 is removed to pay the rent on the space they occupy. What remains is split evenly between the seven growers. How much money is taken in total if each grower receives a $175 share?

Let us translate the text above into an equation. The unknown is going to be the total money taken in dollars. We will call this \begin{align*}x\end{align*}.

8.5% of all the money taken is removed for sales tax”. This means that 91.5% of the money remains. This is \begin{align*}0.915x\end{align*}.

\begin{align*} (0.915x - 150)\end{align*}

“$150 is removed to pay the rent on the space they occupy”

\begin{align*} \frac{0.915x - 150} {7}\end{align*}

“What remains is split evenly between the 7 growers”

If each grower’s share is $175, then we can write the following equation.

\begin{align*} \frac{0.915x -150} {7} & = 175\ && \text{Multiply by both sides}\ 7. \\ 0.915x-150 & = 1225\ && \text{Add}\ 150\ \text{to both sides}. \\ 0.915 x & = 1375\ && \text{Divide by}\ 0.915.\\ \frac{0.915x} {0.915} & = \frac{1373} {0.915} \\ & = 1502.7322 \ldots && \text{Round to two decimal places}.\end{align*}


If the growers are each to receive a $175 share then they must take at least $1,502.73.

Example 6

A factory manager is packing engine components into wooden crates to be shipped on a small truck. The truck is designed to hold sixteen crates, and will safely carry a 1200 lb cargo. Each crate weighs twelve lbs empty. How much weight should the manager instruct the workers to put in each crate in order to get the shipment weight as close as possible to 1200 lbs?

The unknown quantity is the weight to put in each box. This is \begin{align*}x\end{align*}. Each crate, when full will weigh:

\begin{align*}(x + 12) &&& 16\ \text{crates must weigh}. \\ 16(x + 12) &&& \text{And this must equal}\ 16 \ lbs. \\ 16(x + 12) & = 1200\ && \text{Isolate}\ x\ \text{first, divide both sides by}\ 16. \\ x + 12 & = 75\ && \text{Next subtract}\ 12 \ \text{from both sides}. \\ x & = 63\end{align*}


The manager should tell the workers to put 63 lbs of components in each crate.

Ohm’s Law

The electrical current, \begin{align*}I\end{align*} (amps), passing through an electronic component varies directly with the applied voltage, \begin{align*}V\end{align*} (volts), according to the relationship:

\begin{align*}V=I \cdot R\end{align*} where \begin{align*}R\end{align*} is the resistance (measured in Ohms - \begin{align*}\Omega\end{align*})

Example 7

A scientist is trying to deduce the resistance of an unknown component. He labels the resistance of the unknown component \begin{align*}x\Omega\end{align*}. The resistance of a circuit containing a number of these components is \begin{align*}(5x + 20)\Omega\end{align*}. If a 120 volt potential difference across the circuit produces a current of 2.5 amps, calculate the resistance of the unknown component.

Substitute \begin{align*}V=120, I=2.5\end{align*} and \begin{align*}R=5x+20\end{align*} into \begin{align*}V=I \cdot R\end{align*}:

\begin{align*}120 & = 2.5(5x + 20) & & \text{Distribute the} \ 2.5.\\ 120 & = 12.5x \cancel{+50} & & \text{Subtract} \ 50 \ \text{from both sides}.\\ -50 & \qquad \quad \ \ \cancel{-50}\\ 70 & =12.5x & & \text{Divide both sides by} \ 12.5.\\ \frac{70}{12.5}& =\frac{12.5x}{12.5}\\ 5.6\Omega & = x\end{align*}


The unknown components have a resistance of \begin{align*}5.6 \Omega\end{align*}.

Distance, Speed and Time

The speed of a body is the distance it travels per unit of time. We can determine how far an object moves in a certain amount of time by multiplying the speed by the time. Here is our equation.

\begin{align*}\text{distance} = \text{speed} \times \text{time}\end{align*}

Example 8

Shanice’s car is traveling 10 miles per hour slower than twice the speed of Brandon’s car. She covers 93 miles in 1 hour 30 minutes. How fast is Brandon driving?

Here we have two unknowns in this problem. Shanice’s speed and Brandon’s speed. We do know that Shanice’s speed is ten less than twice Brandon’s speed. Since the question is asking for Brandon’s speed, it is his speed in miles per hour that will be \begin{align*}x\end{align*}.

Substituting into the distance time equation yields:

\begin{align*}93& =2x-10 \times 1.5 & & \text{Divide by} \ 1.5.\\ 62& =2x\cancel{-10}\\ +10 & \qquad \ \cancel{+10} & & \text{Add} \ 10 \ \text{to both sides}.\\ 72& =2x\\ \frac{72}{2}&=\frac{2x}{2}& & \text{Divide both sides by} \ 2.\\ 36& =x\end{align*}


Peter is driving at 36 miles per hour.

This example may be checked by considering the situation another way: We can use the fact that Shanice's covers 93 miles in 1 hour 30 minutes to determine her speed (we will call this \begin{align*}y\end{align*} as \begin{align*}x\end{align*} has already been defined as Brandon’s speed):

\begin{align*}93 & = y \cdot 1.5\\ \frac{93} {1.5} & = \frac{1.5y}{1.5}& & \text{Divide both sides by} \ 1.5.\\ y&=62 \ mph\end{align*}

We can then use this information to determine Shanice’s speed by converting the text to an equation.

“Shanice’s car is traveling at 10 miles per hour slower than twice the speed of Peter’s car”

Translates to

\begin{align*}y = 2x -10\end{align*}

It is then a simple matter to substitute in our value in for \begin{align*}y\end{align*} and then solve for \begin{align*}x\end{align*}:

\begin{align*}& \ \ 62 = (2x - \cancel{10})\\ & \underline{+10 \qquad \ \ +\cancel{10}} && \text{Add} \ 10 \ \text{to both sides}.\\ & \ \ 72 = 2x \\ & \ \ 72 = 2x & & \text{Divide both sides by} \ 2.\\ & \ \frac{72} {2} = \frac{2x}{2}\\ & \quad x=36 \ miles \ per \ hour.\end{align*}


Brandon is driving at 36 miles per hour.

You can see that we arrive at exactly the same answer whichever way we solve the problem. In algebra, there is almost always more than one method of solving a problem. If time allows, it is an excellent idea to try to solve the problem using two different methods and thus confirm that you have calculated the answer correctly.

Speed of Sound

The speed of sound in dry air, \begin{align*}v\end{align*}, is given by the following equation.

\begin{align*}v = 331 + 0.6T\end{align*} where \begin{align*}T\end{align*} is the temperature in Celsius and \begin{align*}v\end{align*} is the speed of sound in meters per second.

Example 9

Tashi hits a drainpipe with a hammer and 250 meters away Minh hears the sound and hits his own drainpipe. Unfortunately, there is a one second delay between him hearing the sound and hitting his own pipe. Tashi accurately measures the time from her hitting the pipe and hearing Mihn’s pipe at 2.46 seconds. What is the temperature of the air?

This complex problem must be carefully translated into equations:

\begin{align*}\text{Distance traveled} &= (331 + 0.6T) \times \text{time}\\ \text{time} & = (2.46 - 1) && \text{Do not forget, for one second the sound is not traveling}\\ \text{Distance}& = 2 \times 250\end{align*}

Our equation is:

\begin{align*}2(250) & = (331 + 0.6T)\cdot (2.46 - 1) & & \text{Simplify terms}.\\ \frac{500}{1.46} & = \frac{1.46(331 + 0.6T)}{1.46} & & \text{Divide by} \ 1.46.\\ 342.47 -331 & = 331 + 0.6T -331 & & \text{Subtract 331 from both sides}.\\ \frac{11.47}{0.6} & = \frac{0.6T}{0.6}& & \text{Divide by} \ 0.6.\\ 19.1 & = T\end{align*}


The temperature is 19.1 degrees Celsius.

Lesson Summary

  • If dividing a number outside of parentheses will produce fractions, it is often better to use the Distributive Property (for example, \begin{align*}3(x + 2) = 3x + 6\end{align*}) to expand the terms and then combine like terms to solve the equation.

Review Questions

  1. Solve the following equations for the unknown variable.
    1. \begin{align*} 3 (x - 1) - 2 (x + 3) = 0\end{align*}
    2. \begin{align*} 7(w + 20) - w = 5\end{align*}
    3. \begin{align*} 9(x - 2) = 3x + 3\end{align*}
    4. \begin{align*} 2 \left (5a - \frac{1} {3}\right ) = \frac{2} {7}\end{align*}
    5. \begin{align*} \frac{2} {9} \left (i + \frac{2}{3}\right ) = \frac{2} {5}\end{align*}
    6. \begin{align*} 4 \left (v + \frac{1} {4}\right ) = \frac{35} {2}\end{align*}
    7. \begin{align*} \frac{s - 4} {11} = \frac{2} {5}\end{align*}
    8. \begin{align*} \frac{p} {16} - \frac{2p} {3} = \frac{1} {9}\end{align*}
  2. An engineer is building a suspended platform to raise bags of cement. The platform has a mass of 200 kg, and each bag of cement is 40 kg. He is using two steel cables, each capable of holding 250 kg. Write an equation for the number of bags he can put on the platform at once, and solve it.
  3. A scientist is testing a number of identical components of unknown resistance which he labels \begin{align*}x \Omega\end{align*}. He connects a circuit with resistance \begin{align*}(3x + 4)\Omega\end{align*} to a steady 12 Volt supply and finds that this produces a current of 1.2 Amps. What is the value of the unknown resistance?
  4. Lydia inherited a sum of money. She split it into five equal chunks. She invested three parts of the money in a high interest bank account which added 10% to the value. She placed the rest of her inheritance plus $500 in the stock market but lost 20% on that money. If the two accounts end up with exactly the same amount of money in them, how much did she inherit?
  5. Pang drove to his mother’s house to drop off her new TV. He drove at 50 miles per hour there and back, and spent 10 minutes dropping off the TV. The entire journey took him 94 minutes. How far away does his mother live?

Review Answers

  1. (a) \begin{align*}x = 9\end{align*} (b) \begin{align*}w = -22.5\end{align*} (c) \begin{align*}x = 3.5\end{align*} (d) \begin{align*}a = \frac{2}{21}\end{align*} (e) \begin{align*}i = \frac{17}{15}\end{align*} (f) \begin{align*}v = \frac{33}{8}\end{align*} (g) \begin{align*}s = \frac{42}{5}\end{align*} (h) \begin{align*}p = -\frac{16}{87}\end{align*}
  2. \begin{align*}2(250) = 200 + 40x; x = 7.5 \rightarrow 7\end{align*} bags
  3. \begin{align*}2 \Omega\end{align*}
  4. $1, 176.50
  5. 35 miles

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