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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’ve seen that when we solve for an unknown variable, it can take just one or two steps to get the terms in the right places. Now we’ll 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’ll 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 the variables on the other side. We’ll 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*}.

Before we can combine the variable terms, we need to get rid of that fraction.

First 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\end{align*}.

Combining the fractions then gives us \begin{align*}\frac{3x + 4 - 15x}{3} = 6\end{align*}.

Combining like terms in the numerator gives us \begin{align*}\frac{4 - 12x}{3} = 6\end{align*}.

Multiplying both sides by 3 gives us \begin{align*}4 - 12x = 18\end{align*}.

Subtracting 4 from both sides gives us \begin{align*}-12x = 14\end{align*}.

And finally, dividing both sides by -12 gives us \begin{align*}x = -\frac{14}{12}\end{align*}, which reduces to \begin{align*}x = -\frac{7}{6}\end{align*}.

Solving Multi-Step Equations Using the Distributive Property

You may have noticed that when one side of the equation is multiplied by a constant term, we can either distribute it or just divide it out. If we can divide it out without getting awkward fractions as a result, then that’s usually the better choice, because it gives us smaller numbers to work with. But if dividing would result in messy fractions, then it’s usually better to distribute the constant and go from there.

Example 2

Solve \begin{align*}7(2x - 5) = 21\end{align*}.

The first thing we want to do here is get rid of the parentheses. We could use the Distributive Property, but it just so happens that 7 divides evenly into 21. That suggests that dividing both sides by 7 is the easiest way to solve this problem.

If we do that, we get \begin{align*}2x - 5 = \frac{21}{7} \end{align*} or just \begin{align*}2x - 5 = 3\end{align*}. Then all we need to do is add 5 to both sides to get \begin{align*}2x = 8\end{align*}, and then divide by 2 to get \begin{align*}x = 4\end{align*}.

Example 3

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

Once again, we want to get rid of those parentheses. We could divide both sides by 17, but that would give us an inconvenient fraction on the right-hand side. In this case, distributing is the easier way to go.

Distributing the 17 gives us \begin{align*} 51x + 68 = 7\end{align*}. Then we subtract 68 from both sides to get \begin{align*}51x = -61\end{align*}, and then we divide by 51 to get \begin{align*}x = \frac{-61}{51}\end{align*}. (Yes, that’s a messy fraction too, but since it’s our final answer and we don’t have to do anything else with it, we don’t really care how messy it is.)

Example 4

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

Before we can collect like terms, we need to get rid of the parentheses using the Distributive Property. That gives us \begin{align*}12x - 16 - 14x - 21 = 3\end{align*}, which we can rewrite as \begin{align*}(12x - 14x) + (-16 - 21) = 3\end{align*}. This in turn simplifies to \begin{align*}-2x - 37 = 3\end{align*}.

Next we add 37 to both sides to get \begin{align*}-2x = 40\end{align*}.

And finally, we divide both sides by -2 to get \begin{align*}x = -20\end{align*}.

Example 5

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’s often easier to convert decimals to fractions, but in this equation the fractions are easy to convert to decimals—and with decimals we don’t need to find a common denominator!

In decimal form, our equation becomes \begin{align*}0.1(3.2 + 2x) + 0.5(3 - 0.2x) = 0\end{align*}.

Distributing to get rid of the parentheses, we get \begin{align*}0.32 + 0.2x + 1.5 - 0.1x = 0\end{align*}.

Collecting and combining like terms gives us \begin{align*}0.1x +1.82 = 0\end{align*}.

Then we can subtract 1.82 from both sides to get \begin{align*}0.1x = -1.82\end{align*}, and finally divide by 0.1 (or multiply by 10) to get \begin{align*}x = -18.2\end{align*}.

Solving Real-World Problems Using Multi-Step Equations

Example 6

A growers’ 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 in is set aside for sales tax. $150 goes to pay the rent on the space they occupy. What remains is split evenly between the seven growers. How much total money is taken in if each grower receives a $175 share?

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

“8.5% of all the money taken in is set aside for sales tax." This means that 91.5% of the money remains. This is \begin{align*}0.915x\end{align*}.

“$150 goes to pay the rent on the space they occupy.” This means that what’s left is \begin{align*}0.915x - 150\end{align*}.

“What remains is split evenly between the 7 growers.” That means each grower gets \begin{align*}\frac{0.915x - 150}{7}\end{align*}.

If each grower’s share is $175, then our equation to find \begin{align*}x\end{align*} is \begin{align*}\frac{0.915x - 150}{7} = 175\end{align*}.

First we multiply both sides by 7 to get \begin{align*}0.915x - 150 = 1225\end{align*}.

Then add 150 to both sides to get \begin{align*}0.915x = 1375\end{align*}.

Finally divide by 0.915 to get \begin{align*}x \approx 1502.7322\end{align*}. Since we want our answer in dollars and cents, we round to two decimal places, or $1502.73.

The workers take in a total of $1502.73.

Example 7

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 12 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, so we’ll call that \begin{align*}x\end{align*}.

Each crate when full will weigh \begin{align*}x + 12 \ lbs\end{align*}, so all 16 crates together will weigh \begin{align*}16(x + 12) \ lbs\end{align*}.

We also know that all 16 crates together should weigh 1200 lbs, so we can say that \begin{align*}16(x + 12) = 1200\end{align*}.

To solve this equation, we can start by dividing both sides by 16: \begin{align*}x + 12 = \frac{1200}{16} = 75\end{align*}.

Then subtract 12 from both sides: \begin{align*}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 8

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.


To solve this, we need to start with the equation \begin{align*}V = I \cdot R\end{align*} and substitute in \begin{align*}V = 120, I = 2.5,\end{align*} and \begin{align*}R = 5x + 20\end{align*}. That gives us \begin{align*}120 = 2.5(5x + 20)\end{align*}.

Distribute the 2.5 to get \begin{align*}120 = 12.5x + 50\end{align*}.

Subtract 50 from both sides to get \begin{align*}70 = 12.5x\end{align*}.

Finally, divide by 12.5 to get \begin{align*}5.6 = 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. That means that we can also find out how far an object moves in a certain amount of time if we know its speed: we use the 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 don’t know either Brandon’s speed or Shanice’s, but since the question asks for Brandon’s speed, that’s what we’ll use as our variable \begin{align*}x\end{align*}.

The distance Shanice covers in miles is 93, and the time in hours is 1.5. Her speed is 10 less than twice Brandon’s speed, or \begin{align*}2x - 10\end{align*} miles per hour. Putting those numbers into the equation gives us \begin{align*}93 = 1.5(2x - 10)\end{align*}.

First we distribute, to get \begin{align*}93 = 3x - 15\end{align*}.

Then we add 15 to both sides to get \begin{align*}108 = 3x\end{align*}.

Finally we divide by 3 to get \begin{align*}36 = x\end{align*}.

Brandon is driving at 36 miles per hour.

We can check this answer by considering the situation another way: we can solve for Shanice’s speed instead of Brandon’s and then check that against Brandon’s speed. We’ll use \begin{align*}y\end{align*} for Shanice’s speed since we already used \begin{align*}x\end{align*} for Brandon’s.

The equation for Shanice’s speed is simply \begin{align*}93 = 1.5y\end{align*}. We can divide both sides by 1.5 to get \begin{align*}62 = y\end{align*}, so Shanice is traveling at 62 miles per hour.

The problem tells us that Shanice is traveling 10 mph slower than twice Brandon’s speed; that would mean that 62 is equal to 2 times 36 minus 10. Is that true? Well, 2 times 36 is 72, minus 10 is 62. The answer checks out.

In algebra, there’s almost always more than one method of solving a problem. If time allows, it’s always a good idea to try to solve the problem using two different methods just to confirm that you’ve got the answer right.

Speed of Sound

The speed of sound in dry air, \begin{align*}v\end{align*}, is given by the 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 between her hitting the pipe and hearing Mihn’s pipe at 2.46 seconds. What is the temperature of the air?

This is a complex problem and we need to be careful in writing our equations. First of all, the distance the sound travels is equal to the speed of sound multiplied by the time, and the speed is given by the equation above. So the distance equals \begin{align*}(331 + 0.6T) \times \text{time}\end{align*}, and the time is \begin{align*}2.46 - 1\end{align*} (because for 1 second out of the 2.46 seconds measured, there was no sound actually traveling). We also know that the distance is \begin{align*}250 \times 2\end{align*} (because the sound traveled from Tashi to Minh and back again), so our equation is \begin{align*}250 \times 2 = (331 + 0.6T)(2.46 - 1)\end{align*}, which simplifies to \begin{align*}500 = 1.46(331 + 0.6T)\end{align*}.

Distributing gives us \begin{align*}500 = 483.26 + 0.876T\end{align*}, and subtracting 483.26 from both sides gives us \begin{align*}16.74 = 0.876T\end{align*}. Then we divide by 0.876 to get \begin{align*}T \approx 19.1\end{align*}.

The temperature is about 19.1 degrees Celsius.

Lesson Summary

  • Multi-step equations are slightly more complex than one - and two-step equations, but use the same basic techniques.
  • If dividing a number outside of parentheses will produce fractions, it is often better to use the Distributive Property 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*}3(x + 3) - 2(x - 1) = 0\end{align*}
    3. \begin{align*}7(w + 20) - w = 5\end{align*}
    4. \begin{align*}5(w + 20) - 10w = 5\end{align*}
    5. \begin{align*}9(x - 2) - 3x = 3\end{align*}
    6. \begin{align*}12(t - 5) + 5 = 0\end{align*}
    7. \begin{align*}2(2d + 1) = \frac{2}{3}\end{align*}
    8. \begin{align*}2 \left ( 5a - \frac{1}{3} \right ) = \frac{2}{7}\end{align*}
    9. \begin{align*}\frac{2}{9} \left ( i + \frac{2}{3} \right ) = \frac{2}{5}\end{align*}
    10. \begin{align*}4 \left ( v + \frac{1}{4} \right ) = \frac{35}{2}\end{align*}
    11. \begin{align*}\frac{g}{10} = \frac{6}{3} \end{align*}
    12. \begin{align*}\frac{s - 4}{11} = \frac{2}{5} \end{align*}
    13. \begin{align*}\frac{2k}{7} = \frac{3}{8} \end{align*}
    14. \begin{align*}\frac{7x + 4}{3} = \frac{9}{2} \end{align*}
    15. \begin{align*}\frac{9y - 3}{6} = \frac{5}{2} \end{align*}
    16. \begin{align*}\frac{r}{3} + \frac{r}{2} = 7\end{align*}
    17. \begin{align*} \frac{p}{16} - \frac{2p}{3} = \frac{1}{9} \end{align*}
    18. \begin{align*}\frac{m + 3}{2} - \frac{m}{4} = \frac{1}{3} \end{align*}
    19. \begin{align*}5 \left ( \frac{k}{3} + 2 \right ) = \frac{32}{3}\end{align*}
    20. \begin{align*}\frac{3}{z} = \frac{2}{5}\end{align*}
    21. \begin{align*}\frac{2}{r} + 2 = \frac{10}{3} \end{align*}
    22. \begin{align*}\frac{12}{5} = \frac{3 + z}{z}\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 parts. 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?

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