What if you had two ratios that you knew were equal to one another, like , in which one of the numbers was unknown? How could you solve for *x*? After completing this Concept, you'll be able to write and solve proportions like this one.

### Watch This

CK-12 Foundation: 0311S Proportions (H264)

### Guidance

When two ratios are equal to each other, we call it a proportion. For example, the equation is a proportion. We know it’s true because we can reduce both fractions to .

(Check this yourself to make sure!)

We often use proportions in science and business—for example, when scaling up the size of something. We generally use them to solve for an unknown, so we use algebra and label the unknown variable .

#### Example A

*A small fast food chain operates 60 stores and makes $1.2 million profit every year. How much profit would the chain make if it operated 250 stores?*

**Solution**

First, we need to write a **ratio:** the ratio of profit to number of stores. That would be .

Now we want to know how much profit 250 stores would make. If we label that profit , then the ratio of profit to stores in that case is .

Since we’re assuming the profit is proportional to the number of stores, the ratios are equal and our proportion is .

(Note that we can drop the units – not because they are the same in the numerator and denominator, but because they are the same on both sides of the equation.)

To solve this equation, first we simplify the left-hand fraction to get . Then we multiply both sides by 250 to get .

**If the chain operated 250 stores, the annual profit would be 5 million dollars.**

**Solve Proportions Using Cross Products**

One neat way to simplify proportions is to cross multiply. Consider the following proportion:

If we want to eliminate the fractions, we could multiply both sides by 4 and then multiply both sides by 5. But suppose we just do both at once?

Now comparing this to the proportion we started with, we see that the denominator from the left hand side ends up being multiplied by the numerator on the right hand side. You can also see that the denominator from the *right* hand side ends up multiplying the numerator on the *left* hand side.

In effect the two denominators have multiplied across the equal sign:

becomes .

This movement of denominators is known as **cross multiplying**. It is extremely useful in solving proportions, especially when the unknown variable is in the denominator.

#### Example B

*Solve this proportion for :*

**Solution**

Cross multiply to get , or . Then divide both sides by 4 to get , or .

#### Example C

*Solve the following proportion for :*

**Solution**

Cross multiply to get , or Then divide both sides by 0.5 to get

**Solve Real-World Problems Using Proportions**

#### Example D

*A cross-country train travels at a steady speed. It covers 15 miles in 20 minutes. How far will it travel in 7 hours assuming it continues at the same speed?*

**Solution**

We’ve done speed problems before; remember that speed is just the ratio , so that ratio is the one we’ll use for our proportion. We can see that the speed is , and that speed is also equal to .

To set up a proportion, we first have to get the units the same. 20 minutes is of an hour, so our proportion will be . This is a very awkward looking ratio, but since we’ll be cross multiplying, we can leave it as it is.

Cross multiplying gives us . Multiplying both sides by 3 then gives us , or .

**The train will travel 315 miles in 7 hours.**

#### Example E

*In the United Kingdom, Alzheimer’s disease is said to affect one in fifty people over 65 years of age. If approximately 250000 people over 65 are affected in the UK, how many people over 65 are there in total?*

**Solution**

The fixed ratio in this case is the 1 person in 50. The unknown quantity is the total number of people over 65. Note that in this case we don’t need to include the units, as they will cancel between the numerator and denominator.

Our proportion is . Each ratio represents

.

Cross multiplying, we get , or .

**There are approximately 12.5 million people over the age of 65 in the UK.**

Watch this video for help with the Examples above.

### Guided Practice

*A chemical company makes up batches of copper sulfate solution by adding 250 kg of copper sulfate powder to 1000 liters of water. A laboratory chemist wants to make a solution of identical concentration, but only needs 350 mL (0.35 liters) of solution. How much copper sulfate powder should the chemist add to the water?*

**Solution**

The ratio of powder to water in the first case, in kilograms per liter, is , which reduces to . In the second case, the unknown amount is how much powder to add. If we label that amount , the ratio is . So our proportion is .

To solve for , first we multiply both sides by 0.35 to get , or .

**The mass of copper sulfate that the chemist should add is 0.0875 kg, or 87.5 grams.**

### Explore More

Solve the following proportions.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.9.