# 2.4.6: Applications of Multi-Step Equations

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## Solving Real-World Problems Using Multi-Step Equations

**Real-World Application of Multi-Step Equations **

Use strategies for solving multi-step equations to solve real world equations.

**Real-World Application: Profit **

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 x.

* “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 0.915x.

* “$150 goes to pay the rent on the space they occupy.”* This means that what’s left is 0.915x−150.

* “What remains is split evenly between the 7 growers.”* That means each grower gets (0.915x−150)/7.

If each grower’s share is $175, then our equation to find x is (0.915x−150)/7=175.

First we multiply both sides by 7 to get 0.915x−150=1225.

Then add 150 to both sides to get 0.915x=1375.

Finally divide by 0.915 to get x≈1502.7322. 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.**

**Real-World Application: Ohm’s Law**

**Real-World Application: Ohm’s Law**

The electrical current, I (amps), passing through an electronic component varies directly with the applied voltage, V (volts), according to the relationship V=I⋅R where R is the resistance measured in Ohms (Ω).

A scientist is trying to deduce the resistance of an unknown component. He labels the resistance of the unknown component x Ω. The resistance of a circuit containing a number of these components is (5x+20)Ω. 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 V=I⋅R and substitute in V=120,I=2.5, and R=5x+20. That gives us 120=2.5(5x+20).

Distribute the 2.5 to get 120=12.5x+50.

Subtract 50 from both sides to get 70=12.5x.

Finally, divide by 12.5 to get 5.6=x.

**The unknown components have a resistance of 5.6 Ω.**

**Real-World Application: Distance, Speed and Time**

**Real-World Application: 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 “distance=speed×time.”

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 x.

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 2x−10 miles per hour. Putting those numbers into the equation gives us 93=1.5(2x−10).

First we distribute, to get 93=3x−15.

Then we add 15 to both sides to get 108=3x.

Finally we divide by 3 to get 36=x.

**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 y for Shanice’s speed since we already used x for Brandon’s.

The equation for Shanice’s speed is simply 93=1.5y. We can divide both sides by 1.5 to get 62=y, 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.

**Real-World Application: Speed of Sound**

**Real-World Application: Speed of Sound**

The speed of sound in dry air, v, is given by the equation v=331+0.6T, where T is the temperature in Celsius and v is the speed of sound in meters per second.

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 (331+0.6T)×time, and the time is 2.46−1 (because for 1 second out of the 2.46 seconds measured, there was no sound actually traveling). We also know that the distance is 250×2 (because the sound traveled from Tashi to Minh and back again), so our equation is 250×2=(331+0.6T)(2.46−1), which simplifies to 500=1.46(331+0.6T).

Distributing gives us 500=483.26+0.876T, and subtracting 483.26 from both sides gives us 16.74=0.876T. Then we divide by 0.876 to get T≈19.1.

**The temperature is about 19.1 degrees Celsius.**

**Example**

Example \(\PageIndex{1}\)

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?

**Solution**

The unknown quantity is the weight to put in each box, so we’ll call that x.

Each crate when full will weigh x+12 lbs, so all 16 crates together will weigh 16(x+12) lbs.

We also know that all 16 crates together should weigh 1200 lbs, so we can say that 16(x+12)=1200.

To solve this equation, we can start by dividing both sides by 16: x+12=120016=75.

Then subtract 12 from both sides: x=63.

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

**Review **

For 1-6, solve for the variable in the equation.

- (s−4)/11=2/5
- 2k/7=3/8
- (7x+4)/3=9/2
- (9y−3)/6=5/2
- r/3+r/2=7
- p/16−2p/3=1/9
- 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.
- A scientist is testing a number of identical components of unknown resistance which he labels xΩ. He connects a circuit with resistance (3x+4)Ω 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?
- 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?
- 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)

To view the Review answers, open this PDF file and look for section 3.8.

### Additional Resources

PLIX: Play, Learn, Interact, eXplore: **Bowling Ball Delivery**

Video: **Solving Word Problems Involving Writing Equations - Example 1**

Practice: **Applications of Multi-Step Equations**

Real World: **Physics and Football**