Finding the Missing Quantity
A proportion tells us that two ratios represent the same relationship. When one quantity is unknown, that equality gives us enough information to find it.
For example,
\[ \frac{3}{5}=\frac{x}{20} \]says that the ratio \(3:5\) must remain unchanged when the denominator becomes 20. Since 5 was multiplied by 4, the numerator must also be multiplied by 4:
\[ x=3(4)=12. \]Sometimes the scale factor is easy to see. Other proportions are less obvious. For those, algebra gives us a general method that works consistently.
The Fundamental Property of Proportions
Recall that if
\[ \frac{a}{b}=\frac{c}{d}, \qquad b\neq0,\ d\neq0, \]then the cross-products are equal:
\[ ad=bc. \]This property allows a proportion containing fractions to be rewritten as an equation without those denominators.
For example,
\[ \frac{3}{5}=\frac{x}{20} \]gives
\[ 3(20)=5x. \]Therefore,
\[ 60=5x \]and
\[ x=12. \]A General Method for Solving Proportions
When a proportion contains an unknown quantity, the following process is reliable.
- Write the proportion carefully. Make sure corresponding quantities appear in the same positions.
- Identify any denominator restrictions. A denominator can never equal zero.
- Form the cross-products.
- Solve the resulting equation.
- Check the solution. Substitute it into the original proportion when needed.
The algebra may be simple or advanced, but the basic structure remains the same.
Unknown in the Numerator
Consider
\[ \frac{x}{5}=\frac{12}{15}. \]Form the cross-products:
\[ 15x=5(12). \]Simplify:
\[ 15x=60. \]Divide both sides by 15:
\[ x=4. \]Check:
\[ \frac{4}{5}=\frac{12}{15}. \]Since \(\frac{12}{15}\) simplifies to \(\frac45\), the solution is correct.
Unknown in the Denominator
The unknown does not have to appear in the numerator.
Suppose
\[ \frac{7}{9}=\frac{21}{x}. \]Because \(x\) is a denominator,
\[ x\neq0. \]Form the cross-products:
\[ 7x=9(21). \]Therefore,
\[ 7x=189, \]so
\[ x=27. \]The solution satisfies the denominator restriction and gives
\[ \frac79=\frac{21}{27}. \]Proportions with Algebraic Expressions
A proportion may contain an expression rather than a single unknown number.
Consider
\[ \frac{x+2}{5}=\frac{9}{15}. \]Form the cross-products:
\[ 15(x+2)=5(9). \]Simplify:
\[ 15(x+2)=45. \]Divide by 15:
\[ x+2=3. \]Therefore,
\[ x=1. \]Notice that the entire expression \(x+2\) must be treated as one quantity when forming the cross-product.
Variables in Denominators
When a variable appears in a denominator, begin by identifying values that would make the denominator zero.
Consider
\[ \frac{x}{x+2}=\frac35. \]Since \(x+2\) is a denominator,
\[ x+2\neq0, \]so
\[ x\neq-2. \]Now form the cross-products:
\[ 5x=3(x+2). \]Distribute:
\[ 5x=3x+6. \]Subtract \(3x\) from both sides:
\[ 2x=6. \]Therefore,
\[ x=3. \]Since \(3\neq-2\), the solution is permitted.
Variables on Both Sides
Some proportions produce equations with variable terms on both sides.
Consider
\[ \frac{x+1}{x-2}=\frac23. \]First note the restriction
\[ x\neq2. \]Cross multiply:
\[ 3(x+1)=2(x-2). \]Distribute:
\[ 3x+3=2x-4. \]Subtract \(2x\) from both sides:
\[ x+3=-4. \]Therefore,
\[ x=-7. \]The solution does not violate the restriction \(x\neq2\), so it is valid.
When a Proportion Leads to a Quadratic Equation
Cross multiplication does not always produce a linear equation. If variables appear in the right places on both sides of a proportion, their product may create a squared term.
Consider
\[ \frac{x}{4}=\frac{9}{x}. \]Since \(x\) appears in a denominator,
\[ x\neq0. \]Form the cross-products:
\[ x(x)=4(9). \]Therefore,
\[ x^2=36. \]Taking square roots gives
\[ x=\pm6. \]Both values are algebraically valid because neither value makes a denominator zero.
In a real-world problem, however, the context may impose an additional restriction. If \(x\) represents a length, time, price, or other quantity that cannot be negative, only
\[ x=6 \]would make sense in that context.
Respect Denominator Restrictions
Algebraic manipulation can sometimes produce a value that was never allowed in the original proportion.
Consider
\[ \frac{x+1}{x-1} = \frac{2x}{2x-2}. \]Both denominators become zero when
\[ x=1. \]Therefore,
\[ x\neq1. \]Cross multiplication gives
\[ (x+1)(2x-2)=2x(x-1). \]Expanding both sides:
\[ 2x^2-2=2x^2-2x. \]Subtract \(2x^2\) from both sides:
\[ -2=-2x. \]Therefore,
\[ x=1. \]But \(x=1\) makes the original denominators zero. It is therefore not a valid solution.
In this example, the proportion has no solution.
This is why denominator restrictions should be identified before solving and checked again afterward.
Checking a Solution
A good habit is to substitute the proposed solution into the original proportion.
Suppose we solved
\[ \frac{x}{8}=\frac{15}{20} \]and obtained
\[ x=6. \]Substitute \(6\) into the original equation:
\[ \frac68=\frac{15}{20}. \]Simplify both ratios:
\[ \frac34=\frac34. \]The equality is true, so the solution checks.
Sometimes Algebra Is More Than You Need
Cross multiplication is a powerful general method, but it is not always the fastest way to solve a proportion.
Consider
\[ \frac{4}{7}=\frac{x}{21}. \]Since
\[ 7(3)=21, \]the numerator must scale by the same factor:
\[ 4(3)=12. \]Therefore,
\[ x=12. \]The scale-factor method and the cross-product method are both valid. A strong problem solver chooses the method that makes the structure easiest to see.
Solving Through a Unit Rate
Real-world proportions can sometimes be solved most naturally by finding a unit rate first.
Suppose 5 pounds of rice cost $12.50. At the same price per pound, how much would 8 pounds cost?
First find the cost of one pound:
\[ \frac{\$12.50}{5\text{ lb}} = \$2.50\text{ per lb}. \]Then multiply by 8 pounds:
\[ 8(\$2.50)=\$20. \]The answer is
\[ \boxed{\$20}. \]This produces the same result as solving the proportion
\[ \frac{\$12.50}{5\text{ lb}} = \frac{x}{8\text{ lb}}, \]but the unit-rate approach may reveal the meaning of the relationship more clearly.
Common Errors to Avoid
Reversing Only One Ratio
Corresponding quantities must remain in matching positions.
If one ratio is written as
\[ \frac{\text{miles}}{\text{hours}}, \]then the other should also be written as
\[ \frac{\text{miles}}{\text{hours}}. \]Cross Adding Instead of Cross Multiplying
The fundamental property uses products:
\[ \frac{a}{b}=\frac{c}{d} \quad\Longrightarrow\quad ad=bc. \]It does not say
\[ a+d=b+c. \]Forgetting Parentheses
If an entire expression appears in a numerator or denominator, preserve that expression when cross multiplying.
\[ \frac{x+3}{4}=\frac57 \]becomes
\[ 7(x+3)=20, \]not
\[ 7x+3=20. \]Ignoring Denominator Restrictions
Any value that makes an original denominator zero must be excluded, even if later algebra appears to produce that value as a solution.
The Equation Is Only Part of the Problem
In an applied problem, solving the equation is not the final step. The answer must also make sense in the situation being modeled.
A negative algebraic solution may be perfectly valid in a pure equation but impossible if the variable represents a number of people, a physical length, an elapsed time, or a price.
Good proportional reasoning therefore includes three questions:
- Is the relationship actually proportional?
- Did we solve the proportion correctly?
- Does the solution make sense in context?
Check Your Understanding
- Solve: \[ \frac{x}{6}=\frac{10}{15}. \]
- Solve: \[ \frac{8}{12}=\frac{14}{x}. \]
- Solve: \[ \frac{x+4}{6}=\frac{5}{10}. \]
- Solve and state the denominator restriction: \[ \frac{x}{x+1}=\frac23. \]
- Solve: \[ \frac{x}{5}=\frac{20}{x}. \]
- Six notebooks cost $15. At the same unit price, how much should 14 notebooks cost?
Show Answers
- \[ 15x=60 \] so \[ x=4. \]
- \[ 8x=12(14)=168 \] so \[ x=21. \]
- \[ 10(x+4)=30 \] so \[ x+4=3 \] and \[ x=-1. \]
- The restriction is \[ x\neq-1. \] Then \[ 3x=2(x+1), \] so \[ 3x=2x+2 \] and \[ x=2. \]
- Since \(x\neq0\), \[ x^2=100. \] Therefore, \[ x=\pm10. \]
- First find the unit price: \[ \frac{\$15}{6}=\$2.50 \] per notebook. Then \[ 14(\$2.50)=\$35. \]
Key Ideas
- A proportion is an equation stating that two ratios are equal.
- If \(\frac{a}{b}=\frac{c}{d}\), then \(ad=bc\).
- Cross-products convert many proportions into familiar algebraic equations.
- Expressions in a numerator or denominator must be treated as complete quantities.
- Variables in denominators create restrictions that must be identified before solving.
- Some proportions lead to linear equations, while others can lead to quadratic equations.
- A scale factor or unit rate may sometimes solve a proportion more efficiently than cross multiplication.
- Solutions should be checked against both the original equation and the real-world context.