4.1: Proportions

In our discussion of similar triangles the idea of a proportion will play an important role. In this section we will review the important properties of proportions.

A proportion is an equation which states that two fractions are equal. For example, \(\dfrac{2}{6}=\dfrac{4}{12}\) is a proportion. We sometimes say "2 is to 6 as 4 is to \(12\)." This is also written \(2: 6 = 4:12\). The extremes of this proportion are the numbers 2 and 12 and the means are the numbers 6 and 4. Notice that the product of the means \(6 \times 4=24\) is the same as the product of the extremes \(2 \times 12=24\).

Theorem \(\PageIndex{1}\)

If \(\dfrac{a}{b} = \dfrac{c}{d}\) then \(a d= be\). Conversely, if \(ad = bc\) then \(\dfrac{a}{b} = \dfrac{c}{d}.\) (The product of the means is equal to the product of the extremes).

EXAMPLES:

• \(\dfrac{2}{6} = \dfrac{4}{12}\) and \(2 \times 12=6 \times 4\) are both true.
• \(\dfrac{2}{3} = \dfrac{6}{9}\) and \(2 \times 9=3 \times 6\) are both true.
• \(\dfrac{1}{4} = \dfrac{4}{12}\) and \(1 \times 12=4 \times 4\) are both false.

Proof of Theorem 1: If \(\dfrac{a}{b} = \dfrac{c}{d}\), multiply both sides of the equation by \(bd\):

\[\dfrac{a}{\cancel{b}} (\cancel{b} d) = \dfrac{c}{\cancel{d}} (b \cancel{d})\]

We obtain \(ad = bc\).

Conversely, if \(ad = bc\), divide both sides of the equation by \(bd\):

\[\dfrac{d\cancel{d}}{b\cancel{d}} = \dfrac{\cancel{b}c}{\cancel{b}d}\]

The result is \(\dfrac{a}{b} = \dfrac{c}{d}\).

The following theorem shows that we can interchange the means or the extremes or both of them simultaneously and still have a valid proportion:

Theorem \(\PageIndex{2}\)

If one of the following is true then they are all true:

1. \(\dfrac{a}{b}=\dfrac{c}{d}\)
2. \(\dfrac{a}{c}=\dfrac{b}{d}\)
3. \(\dfrac{d}{b}=\dfrac{c}{a}\)
4. \(\dfrac{d}{c}=\dfrac{b}{a}\)

Proof

If any one of these proportions is true when \(ad = bc\) by Theorem \(\PageIndex{1}\). The remaining proportions can then be obtained from \(ad = bc\) by division, as in Theorem \(\PageIndex{1}\).

EXAMPLE: \(\dfrac{2}{6} = \dfrac{4}{12}, \dfrac{2}{4} = \dfrac{6}{12}, \dfrac{12}{6} = \dfrac{4}{2}, \dfrac{12}{4} = \dfrac{6}{2}\) are all true because \(2 \times 12=6 \times 4\).

The process of converting a proportion \(\dfrac{2}{6} = \dfrac{4}{12}\) to the equivalent equation \(2 \times 12 = 6 \times 4\) is sometimes called cross multiplication. The idea is conveyed by the following notation:

Example \(\PageIndex{1}\)

Find \(x: \dfrac{3}{x} = \dfrac{4}{20}\)

Solution

By "cross multiplication,"

\[\begin{array} {rcl} {3(20)} & = & {x(4)} \\ {60} & = & {4x} \\ {15} & = & {x} \end{array}\]

Check:

\(\dfrac{3}{x} = \dfrac{3}{15} = \dfrac{1}{5}\). \(\dfrac{4}{20} = \dfrac{1}{5}\).

Answer: \(x = 15\).

Example \(\PageIndex{2}\)

Find \(x\): \(\dfrac{x - 1}{x - 3} = \dfrac{2x + 2}{x + 1}\)

Solution

\[\begin{array} {rcl} {(x - 1)(x + 1)} & = & {(x - 3)(2x + 2)} \\ {x^2 - 1} & = & {2x^2 - 4x - 6} \\ {0} & = & {x^2 - 4x - 5} \\ {0} & = & {(x - 5)(x + 1)} \\ {0} & = & {x - 5\ \ \ \ \ \ 0 = x + 1} \\ {5} & = & {x \ \ \ \ \ \ \ \ -1 = x} \end{array}\]

Check, \(x = 5\):

\(\dfrac{x - 1}{x - 3} = \dfrac{5 - 1}{5 - 3} = \dfrac{4}{2} = 2\). \(\dfrac{2x + 2}{x + 1} = \dfrac{2(5) + 2}{5 + 1} = \dfrac{12}{6} = 2\).

Check, \(x = -1\):

\(\dfrac{x - 1}{x - 3} = \dfrac{-1 -1}{-1 - 3} = \dfrac{-2}{-4} = \dfrac{1}{2}\). \(\dfrac{2x + 2}{x + 1} = \dfrac{2(-1) + 2}{-1 + 1} = \dfrac{-2 + 2}{0} = \dfrac{0}{0}\).

Since \(\dfrac{0}{0}\) is undefined, we reject this answer.

Answer: \(x = 5\).

Problems

1 - 12. Find \(x\):

1. \(\dfrac{6}{x} = \dfrac{18}{3}\)

2. \(\dfrac{4}{x} = \dfrac{2}{6}\)

3. \(\dfrac{7}{1} = \dfrac{x}{3}\)

4. \(\dfrac{x}{8} = \dfrac{9}{6}\)

5. \(\dfrac{7}{1} = \dfrac{x}{3}\)

6. \(\dfrac{10}{2} = \dfrac{25}{x}\)

7. \(\dfrac{x + 5}{x} = \dfrac{5}{4}\)

8. \(\dfrac{x - 6}{4} = \dfrac{5}{10}\)

9. \(\dfrac{3 + x}{x} = \dfrac{3}{2}\)

10. \(\dfrac{x}{x+3} = \dfrac{4}{x}\)

11. \(\dfrac{3x - 3}{2x + 6} = \dfrac{x - 1}{x}\)

12. \(\dfrac{3x - 6}{x - 2} = \dfrac{2x + 2}{x - 1}\)