Chapter 2 ยท Proportional reasoning foundations
Solve, Interpret, and Check Proportions
Solve proportions with cross products, set them up for constant-rate word problems, and find and correct errors in worked solutions.
You already know how to find a unit rate, such as the number of liters a pump moves in minute. Now you will use equal ratios to find a number you do not know yet.
A ratio compares two quantities by division, and you can write it as a fraction. A proportion is an equation that says two ratios are equal. Every denominator in a proportion must be nonzero, because division by zero is undefined. Here is a proportion:
This proportion is true because both ratios have the same value: and . The divisors and are nonzero, so both quotients exist. You can also see it by scaling: multiplying and by the same nonzero number, , gives and without changing the value of the ratio.
To solve a proportion means to find the value of the unknown that makes the two ratios equal. Take , where stands for an unknown number and both denominators are nonzero. The denominator is multiplied by to make , so the numerator must be multiplied by as well: .
Scaling is quick when the multiplier is easy to spot. When it is not, you need a method that does not depend on spotting it.
Let , , and stand for numbers, with and , and suppose .
Multiply both sides by . Because and are both nonzero, is nonzero too. Multiplying both sides of an equation by the same nonzero number, positive or negative, gives an equation with exactly the same solutions.
On the left, divides out and leaves . On the right, divides out and leaves . The products and are called cross products, because each numerator is multiplied by the denominator of the other ratio.
Key concept
Equal ratios have equal cross products
When and , the proportion is true exactly when the cross products and are equal.
To solve a proportion:
- Note which values the unknown may take. An unknown in a denominator cannot be , because division by zero is undefined.
- Write the cross-product equation.
- If the number multiplying the unknown is nonzero, divide both sides by it.
- Check your value in the original proportion.
Worked example
Unknown in a numerator
Solve , where stands for an unknown number.
Check the conditions. The denominators 15 and 6 are both nonzero, so both ratios are defined. The unknown x is a numerator, so it may be any number.
Write the cross products: multiply each numerator by the other ratio's denominator. This is the same as multiplying both sides by 15 times 6, which is 90 and not zero, so the new equation has the same solution.
Multiply on each side: x times 6 is 6x, and 15 times 4 is 60.
Divide both sides by 6. Dividing both sides by the same nonzero number keeps the equation true for the same x.
Check in the original proportion. Put 10 in place of x and simplify each ratio. Both ratios equal two thirds, so they match.
Answer
The solution is , because and both equal .
Worked example
Unknown in a denominator
Solve , where stands for an unknown nonzero number.
State the condition on the unknown. Here x is a denominator, and division by zero is undefined, so x cannot be 0. The other denominator, 5, is also nonzero.
Write the cross products. This multiplies both sides by 5x, which is nonzero because 5 and x are both nonzero, so the new equation has the same solution.
Multiply on each side: 14 times 5 is 70, and x times 4 is 4x.
Divide both sides by 4, which is nonzero, to leave x alone.
Check. The value 17.5 is not zero, so it is allowed as a denominator. Put it in place of x: 14 divided by 17.5 is 0.8, and 4 divided by 5 is 0.8.
Answer
The solution is . This value is exact, and it makes both ratios equal to .
A proportion fits a word problem only when the two quantities are in a proportional relationship: every pair of matching values gives the same ratio, so the unit rate is constant. The quantity you divide by must be nonzero. Look for words such as at a constant rate or at the same rate. If the rate changes, a proportion does not apply.
When it does apply:
- Define the variable. Say what it counts, its unit, and which values make sense.
- Write both ratios with the same units in the same positions.
- Solve, then check the answer against the situation.
The table organizes a problem about a pump that runs at a constant rate. It moves liters in minutes, and you want the number of minutes, , that it needs to move liters. A pumping time is positive, so .
| Quantity | Known run | New run |
|---|---|---|
Volume (liters) | ||
Time (minutes) |
Worked example
A constant-rate word problem
A pump fills a tank at a constant rate. It moves liters in minutes. How many minutes does it need to move liters?
Define the variable and its domain. Let t be the number of minutes the pump needs to move 99 liters. Pumping takes a positive amount of time, so t is greater than 0. That means t is not zero and may be a denominator.
Read each column of the table as a ratio of liters to minutes. The rate is constant, so the two ratios are equal. Liters are on top in both ratios, and minutes are on the bottom in both.
Write the cross products. The denominators 8 and t are both nonzero, so this step keeps the same solution.
Multiply on the right: 8 times 99 is 792.
Divide both sides by 36, which is nonzero.
Check with unit rates and with size. The divisors 8 and 22 are nonzero, and both runs give 4.5 liters per minute. Also, 99 liters is between 2 and 3 times 36 liters, and 22 minutes is between 16 and 24 minutes.
Answer
The pump needs minutes to move liters.
Checking also helps you judge a finished solution, including your own. Errors can hide in the setup, as in the warning above, or in the solving steps. To find and correct an error:
- Test the answer. Put it into the original proportion. Use the exact value, because a rounded value is only an approximation and can make equal ratios look slightly different. If the two ratios are not equal, there is an error.
- Trace the lines. For each line, ask what was done to both sides and whether that step keeps the equation true.
- Fix the first wrong line, then finish and check again.
Dev solved , where stands for an unknown nonzero number. His work is shown below.
| Line | Dev wrote |
|---|---|
1 | |
2 | |
3 |
Worked example
Find and correct the error
Dev solved , where stands for an unknown nonzero number. He got . Decide whether his answer is correct. If it is not, find the first wrong line in his work and correct it.
Test the answer with its exact value. First simplify 96/21 to 32/7 by dividing the top and bottom by 3. Then put 32/7 in place of x on the left side. Dividing by the nonzero fraction 32/7 is the same as multiplying by its reciprocal, 7/32.
Compare with the right side. 12/21 simplifies to 4/7 when you divide the top and bottom by 3. The left side, 7/4, is greater than 1, and the right side, 4/7, is less than 1. The ratios are not equal, so the answer is wrong.
Trace the lines. A valid first step multiplies both sides by 21x, which is nonzero. That gives 8 times 21 on the left and 12 times x on the right. Line 1 instead multiplies the two numerators together and the two denominators together, so it does not follow from the proportion. Lines 2 and 3 follow correctly from line 1, so line 1 is the only error.
Correct line 1. Both denominators, x and 21, are nonzero, so write the true cross products: 8 times 21 and x times 12.
Multiply on each side, then divide both sides by 12, which is nonzero.
Check the corrected answer. The value 14 is not zero, so it is allowed as a denominator. Put it in place of x: both ratios simplify to 4/7.
Answer
Dev's answer is not correct. His error is in line 1, where he multiplied straight across instead of forming cross products. The correct solution is .
Takeaway. A proportion says that two ratios are equal, and every denominator in it must be nonzero. Under that condition the cross products are equal, so you can rewrite the proportion as a simpler equation and divide both sides by the number multiplying the unknown, as long as that number is not zero. In a word problem, use a proportion only when the rate is constant, define your variable with its unit, and keep matching units in matching positions. Always test an answer in the original proportion with its exact value. That test is also how you find errors in a solution.
Next, you will use proportions to solve word problems about scale drawings.