Math Core

If 3 notebooks cost $7.50, what do 8 cost? When two ratios are equal, you can write an equation called a proportion and solve it for the missing number. This is one of your first real equations in algebra, and the skills you use here (multiplying both sides, isolating a variable) will come up again and again.

What a proportion is

Definition

Proportion

A proportion is an equation stating that two ratios are equal:

\frac{a}{b} = \frac{c}{d}

For example, $\dfrac{4}{6} = \dfrac{10}{15}$ is a true proportion, because both fractions simplify to $\dfrac{2}{3}$ .

Cross products

Start with $\dfrac{a}{b} = \dfrac{c}{d}$ and multiply both sides by $bd$ to clear the fractions:

\begin{aligned} bd \cdot \frac{a}{b} &= bd \cdot \frac{c}{d} \\ ad &= bc \end{aligned}

The $b$ cancels on the left and the $d$ cancels on the right. That result is so useful it has a name.

The cross products property

If $\dfrac{a}{b} = \dfrac{c}{d}$ (with $b$ and $d$ not zero), then

ad = bc.

The cross products are equal. This works in reverse too: if $ad = bc$ , the two ratios are equal.

Check $\dfrac{4}{6} = \dfrac{10}{15}$ : $4 \times 15 = 60$ and $6 \times 10 = 60$ . The cross products match, so it's a proportion.

Solving a proportion

When one of the four numbers is unknown, call it $x$ , set the cross products equal, and solve the equation you get.

Worked example: Unknown in the numerator

Solve $\dfrac{x}{8} = \dfrac{15}{20}$ .

\begin{aligned} 20 \cdot x &= 8 \cdot 15 && \text{cross products} \\ 20x &= 120 \\ x &= \frac{120}{20} = 6 && \text{divide both sides by 20} \end{aligned}

Check: $\dfrac{6}{8} = \dfrac{3}{4}$ and $\dfrac{15}{20} = \dfrac{3}{4}$ . ✓

The unknown can be anywhere in the proportion. The steps are the same.

Worked example: Unknown in the denominator

Solve $\dfrac{5}{x} = \dfrac{20}{36}$ .

\begin{aligned} 5 \cdot 36 &= 20 \cdot x \\ 180 &= 20x \\ 9 &= x \end{aligned}

Check: $\dfrac{5}{9}$ and $\dfrac{20}{36}$ both simplify to $\dfrac{5}{9}$ . ✓

Tip

Sometimes you can solve by scaling instead. In $\dfrac{x}{8} = \dfrac{3}{4}$ , the denominator 4 was multiplied by 2 to get 8, so $x = 3 \times 2 = 6$ . Scaling is fast when the numbers are friendly; cross products always work.

Setting up word problems

To turn a situation into a proportion, write the same comparison on both sides. If the left side is $\dfrac{\text{notebooks}}{\text{dollars}}$ , the right side must also be $\dfrac{\text{notebooks}}{\text{dollars}}$ .

Worked example: A scale map

On a map, 2 cm represents 15 km. Two towns are 7 cm apart on the map. How far apart are they really?

Write $\dfrac{\text{map cm}}{\text{real km}}$ on both sides:

\begin{aligned} \frac{2}{15} &= \frac{7}{x} \\ 2x &= 105 \\ x &= 52.5 \end{aligned}

The towns are 52.5 km apart.

Common mistake

Don't flip one ratio. Writing $\dfrac{2}{15} = \dfrac{x}{7}$ puts real km on top of the right side but map cm on top of the left side, and gives $x = \dfrac{14}{15}$ km, which is far too small. Label each part with its unit before you solve.

When the unknown is inside an expression

In algebra, you'll see proportions where a numerator is an expression like $x + 2$ . Cross products still work. Put parentheses around the expression so the whole thing gets multiplied.

Worked example: An expression in the proportion

Solve $\dfrac{x + 2}{6} = \dfrac{5}{3}$ .

\begin{aligned} 3(x + 2) &= 6 \cdot 5 && \text{cross products} \\ 3(x + 2) &= 30 \\ x + 2 &= 10 && \text{divide both sides by 3} \\ x &= 8 && \text{subtract 2 from both sides} \end{aligned}

Check: $\dfrac{8 + 2}{6} = \dfrac{10}{6} = \dfrac{5}{3}$ . ✓

Practice

Practice 1

Is $\dfrac{6}{9} = \dfrac{8}{12}$ a true proportion?

Practice 2

Solve $\dfrac{x}{5} = \dfrac{12}{20}$ .