The Central Idea
Consider
\[ \frac{3}{5}=\frac{x}{20}. \]There are several legitimate ways to solve this proportion.
We can recognize a scale factor:
\[ 5\cdot4=20, \]so
\[ x=3\cdot4=12. \]We can use cross products:
\[ 3(20)=5x, \] \[ 60=5x, \] \[ x=12. \]Or we can compare unit rates.
The methods look different, but they all preserve the same proportional relationship.
Start by Reading the Proportion
Before manipulating symbols, ask what the proportion says.
For
\[ \frac{3}{5}=\frac{x}{20}, \]the ratio \(3:5\) must match the ratio \(x:20\).
The unknown is not floating in isolation. It belongs to a multiplicative relationship.
Method 1: Use an Obvious Scale Factor
Consider
\[ \frac{4}{7}=\frac{x}{35}. \]The denominator changes from \(7\) to \(35\):
\[ 7\cdot5=35. \]Apply the same scale factor to the numerator:
\[ 4\cdot5=20. \]Therefore,
\[ \boxed{x=20}. \]Scaling Down Works Too
Consider
\[ \frac{18}{24}=\frac{x}{8}. \]The denominator changes from \(24\) to \(8\):
\[ 24\div3=8. \]So divide the numerator by \(3\):
\[ 18\div3=6. \]Thus,
\[ \boxed{x=6}. \]Method 2: Find a Unit Rate
Suppose
\[ \frac{\$15}{5\text{ items}} = \frac{x}{8\text{ items}}. \]First reduce the first rate to one item:
\[ \frac{15}{5}=3. \]So the cost is
\[ \$3\text{ per item}. \]For 8 items:
\[ x=3(8)=24. \]Therefore,
\[ \boxed{x=24}. \]Method 3: Use Equivalent Fractions
Consider
\[ \frac{x}{18}=\frac{5}{6}. \]Since
\[ 6\cdot3=18, \]the equivalent numerator is
\[ 5\cdot3=15. \]Thus,
\[ \boxed{x=15}. \]Method 4: Use Cross Products
For a general proportion
\[ \frac{a}{b}=\frac{c}{d}, \]with
\[ b\ne0,\qquad d\ne0, \]multiply both sides by \(bd\):
\[ ad=bc. \]This is the cross-product relationship.
Cross Multiplication Comes from Equality
Suppose
\[ \frac{a}{b}=\frac{c}{d}. \]Multiply both sides by \(b\):
\[ a=\frac{bc}{d}. \]Then multiply by \(d\):
\[ ad=bc. \]Cross multiplication is simply a compact way to clear the denominators.
Cross-Product Example
Solve
\[ \frac{7}{12}=\frac{x}{30}. \]Cross multiply:
\[ 7(30)=12x. \]So
\[ 210=12x. \]Divide:
\[ x=\frac{210}{12}. \]Simplify:
\[ \boxed{x=\frac{35}{2}=17.5}. \]Choose the Simplest Method
For
\[ \frac{3}{4}=\frac{x}{20}, \]cross multiplication works:
\[ 3(20)=4x. \]But scaling is faster because
\[ 4\cdot5=20. \]Then immediately
\[ x=15. \]Good algebra is not about using the most formal-looking method. It is about seeing structure and choosing an efficient valid method.
Unknown in the First Numerator
Solve
\[ \frac{x}{9}=\frac{8}{12}. \]Cross multiply:
\[ 12x=72. \]Thus,
\[ \boxed{x=6}. \]Unknown in the Second Numerator
Solve
\[ \frac{5}{8}=\frac{x}{24}. \]Since
\[ 8\cdot3=24, \]we get
\[ x=5\cdot3=15. \]Therefore,
\[ \boxed{x=15}. \]Unknown in the First Denominator
Solve
\[ \frac{6}{x}=\frac{9}{15}. \]Domain restriction:
\[ x\ne0. \]Cross multiply:
\[ 6(15)=9x. \]So
\[ 90=9x, \] \[ \boxed{x=10}. \]Unknown in the Second Denominator
Solve
\[ \frac{4}{7}=\frac{12}{x}. \]Domain:
\[ x\ne0. \]Cross multiply:
\[ 4x=84. \]Therefore,
\[ \boxed{x=21}. \]Variables in More Than One Place
Consider
\[ \frac{x}{5}=\frac{x+6}{8}. \]Cross multiply:
\[ 8x=5(x+6). \]Distribute:
\[ 8x=5x+30. \]Subtract \(5x\):
\[ 3x=30. \]So
\[ \boxed{x=10}. \]Variable Expressions in Denominators
Consider
\[ \frac{2}{x-1}=\frac{5}{x+2}. \]First record the domain restrictions:
\[ x\ne1, \] \[ x\ne-2. \]Cross multiply:
\[ 2(x+2)=5(x-1). \]Expand:
\[ 2x+4=5x-5. \]So
\[ 9=3x, \] \[ x=3. \]The candidate satisfies both restrictions, so
\[ \boxed{x=3}. \]Domain Restrictions Come Before Simplification
Suppose
\[ \frac{x-2}{x-2}=\frac{3}{3}. \]The left side requires
\[ x\ne2. \]For allowed values, both sides equal \(1\).
So every real number except \(2\) satisfies the proportion:
\[ \boxed{\mathbb{R}\setminus\{2\}}. \]Cancellation does not erase the original restriction.
A Candidate Can Be Excluded
Consider
\[ \frac{x+1}{x-3}=\frac{4}{x-3}. \]Domain:
\[ x\ne3. \]Clearing the common denominator gives
\[ x+1=4. \]So
\[ x=3. \]But \(x=3\) is outside the original domain.
Therefore,
\[ \boxed{\varnothing}. \]Proportions with Fractions Inside Them
Solve
\[ \frac{\frac23}{5} = \frac{x}{15}. \]Simplify the left side:
\[ \frac{\frac23}{5} = \frac23\cdot\frac15 = \frac{2}{15}. \]So
\[ \frac{2}{15} = \frac{x}{15}. \]Therefore,
\[ \boxed{x=2}. \]Complex Proportion Example
Solve
\[ \frac{x/4}{3} = \frac{5}{6}. \]Simplify the left side:
\[ \frac{x/4}{3} = \frac{x}{12}. \]So
\[ \frac{x}{12} = \frac56. \]Cross multiply:
\[ 6x=60. \]Thus,
\[ \boxed{x=10}. \]Reduce Before Cross Multiplying When Helpful
Consider
\[ \frac{18}{24}=\frac{x}{40}. \]Reduce the first ratio:
\[ \frac{18}{24}=\frac34. \]Now solve
\[ \frac34=\frac{x}{40}. \]Since
\[ 4\cdot10=40, \]we get
\[ \boxed{x=30}. \]Simplifying first can make the structure obvious.
Cancel Before Multiplying Large Numbers
Suppose
\[ \frac{24}{35}=\frac{x}{70}. \]Because
\[ 35\cdot2=70, \]we immediately get
\[ x=48. \]This avoids creating unnecessarily large cross products.
Decimal Proportions
Solve
\[ \frac{2.5}{4}=\frac{x}{10}. \]Cross multiply:
\[ 4x=25. \]So
\[ \boxed{x=6.25}. \]Convert Decimals to Fractions When Useful
Consider
\[ \frac{0.75}{3}=\frac{x}{8}. \]Since
\[ 0.75=\frac34, \]the left side is
\[ \frac{\frac34}{3} = \frac14. \]Thus,
\[ \frac14=\frac{x}{8}. \]So
\[ \boxed{x=2}. \]Proportions with Units
Suppose 5 pounds cost \$13.75. What do 8 pounds cost at the same rate?
Write
\[ \frac{\$13.75}{5\text{ lb}} = \frac{x}{8\text{ lb}}. \]Cross multiply:
\[ 5x=13.75(8). \]So
\[ x=22. \]Therefore,
\[ \boxed{\$22}. \]Use Units to Check the Setup
If the proportion is
\[ \frac{\text{dollars}}{\text{pounds}} = \frac{\text{dollars}}{\text{pounds}}, \]the units align correctly.
If instead one side is
\[ \frac{\text{dollars}}{\text{pounds}} \]and the other is
\[ \frac{\text{pounds}}{\text{dollars}}, \]the setup compares reciprocal rates and is probably not the intended proportion.
Solving Percent Proportions
Find \(x\) if \(30\%\) of \(80\) is \(x\).
Use
\[ \frac{30}{100}=\frac{x}{80}. \]Cross multiply:
\[ 100x=2400. \]Therefore,
\[ \boxed{x=24}. \]Solving Scale Problems
A map uses
\[ 1\text{ cm}:6\text{ km}. \]If two cities are 8 cm apart on the map, let \(d\) be the actual distance.
\[ \frac{1\text{ cm}}{6\text{ km}} = \frac{8\text{ cm}}{d}. \]Cross multiply:
\[ d=48. \]So the real distance is
\[ \boxed{48\text{ km}}. \]Solving Similar-Figure Proportions
Suppose corresponding sides satisfy
\[ \frac{6}{9}=\frac{14}{x}. \]Cross multiply:
\[ 6x=126. \]Thus,
\[ \boxed{x=21}. \]Solving Rate Proportions
A machine makes 90 parts in 3 hours. How many parts does it make in 7 hours at the same rate?
Write
\[ \frac{90\text{ parts}}{3\text{ hr}} = \frac{x\text{ parts}}{7\text{ hr}}. \]The unit rate is
\[ 30\text{ parts per hour}. \]So
\[ x=30(7)=210. \]Therefore,
\[ \boxed{210\text{ parts}}. \]Cross Multiplication Is Often Algebraically Equivalent to Unit-Rate Reasoning
For
\[ \frac{90}{3}=\frac{x}{7}, \]cross multiplication gives
\[ 3x=630, \] \[ x=210. \]Unit-rate reasoning gives
\[ 90\div3=30, \] \[ 30\cdot7=210. \]The same proportional structure supports both methods.
When There Is No Constant Proportion
Suppose a taxi charges
\[ C=3m+5. \]It would be incorrect to use
\[ \frac{C_1}{m_1} = \frac{C_2}{m_2} \]because the fixed \$5 fee prevents the cost-to-mile ratio from remaining constant.
Do not solve with proportions unless the underlying relationship is actually proportional.
Check for Proportional Structure First
Before writing a proportion, ask:
- Does one quantity scale by the same factor as the other?
- Is there a constant unit rate?
- Would doubling one quantity double the other?
- Would zero of one quantity correspond to zero of the other?
If not, a proportion may be the wrong model.
Verification by Substitution
Suppose we solve
\[ \frac{4}{7}=\frac{x}{21} \]and obtain
\[ x=12. \]Substitute:
\[ \frac47=\frac{12}{21}. \]Simplify:
\[ \frac{12}{21}=\frac47. \]The candidate is verified.
Verification by Cross Products
For
\[ \frac47=\frac{12}{21}, \]check:
\[ 4(21)=84, \] \[ 7(12)=84. \]Equal cross products confirm the proportion.
Estimate Before Solving
Suppose
\[ \frac{5}{8}=\frac{x}{40}. \]Since \(40\) is five times \(8\), we should expect \(x\) to be five times \(5\):
\[ x\approx25. \]The exact solution
\[ x=25 \]matches the estimate.
Estimating helps catch reversed ratios and arithmetic mistakes.
Worked Example: Variable in Both Ratios
Solve
\[ \frac{x+1}{6}=\frac{x-2}{3}. \]Cross multiply:
\[ 3(x+1)=6(x-2). \]Expand:
\[ 3x+3=6x-12. \]Add \(12\):
\[ 3x+15=6x. \]Subtract \(3x\):
\[ 15=3x. \]Thus,
\[ \boxed{x=5}. \]Worked Example: Variable Denominators
Solve
\[ \frac{3}{x+1}=\frac{6}{x+5}. \]Restrictions:
\[ x\ne-1, \qquad x\ne-5. \]Cross multiply:
\[ 3(x+5)=6(x+1). \]Expand:
\[ 3x+15=6x+6. \]Then
\[ 9=3x, \] \[ x=3. \]The candidate is allowed, so
\[ \boxed{x=3}. \]Worked Example: A Restricted Identity
Consider
\[ \frac{x+2}{x-1} = \frac{2x+4}{2x-2}. \]Restrictions:
\[ x\ne1. \]Factor the right side:
\[ \frac{2(x+2)}{2(x-1)} = \frac{x+2}{x-1}. \]So the proportion is true for every allowed \(x\).
Solution set:
\[ \boxed{\mathbb{R}\setminus\{1\}}. \]A Reliable Solving Workflow
- Confirm that the relationship is proportional.
- Align corresponding quantities consistently.
- Record all denominator restrictions.
- Look for a simple scale factor.
- Consider using a unit rate.
- Simplify ratios before doing heavier algebra when useful.
- Use cross products when no simpler structure is obvious.
- Solve the resulting equation carefully.
- Reject candidates outside the original domain.
- Verify the result in the original proportion and interpret it in context.
Common Errors
Cross Multiplying Before Setting Up Corresponding Ratios Correctly
Perfect algebra cannot repair an incorrectly structured proportion.
Automatically Using Cross Multiplication
Scaling or unit-rate reasoning may be faster and clearer.
Forgetting Domain Restrictions
Variable denominators may exclude values that later appear as algebraic candidates.
Cross Multiplying an Expression That Is Not Actually a Proportion
Cross products apply to an equation of two ratios, not to arbitrary sums of fractions.
Using a Proportion for a Nonproportional Model
Fixed fees, changing rates, and additive offsets can invalidate proportional reasoning.
Dropping Units During Setup
Units often reveal whether corresponding quantities are aligned correctly.
Failing to Verify
Substitute the result into the original proportion, especially when variable denominators are present.
Practice
1. Solve by Scaling
Solve:
\[ \frac{3}{8}=\frac{x}{32}. \]Check Answer
Since \(8\cdot4=32\),
\[ x=3\cdot4=12. \]Answer:
\[ \boxed{x=12}. \]2. Solve by Cross Products
Solve:
\[ \frac{7}{9}=\frac{21}{x}. \]Check Answer
\[ 7x=189, \] \[ \boxed{x=27}. \]3. Variable in a Denominator
Solve:
\[ \frac{4}{x}=\frac{10}{15}. \]Check Answer
Restriction:
\[ x\ne0. \]Cross multiply:
\[ 60=10x, \] \[ \boxed{x=6}. \]4. Variable Expressions
Solve:
\[ \frac{x}{4}=\frac{x+3}{7}. \]Check Answer
\[ 7x=4x+12, \] \[ 3x=12, \] \[ \boxed{x=4}. \]5. Unit Rate
Eight pounds of fruit cost \$18.40. What will 12 pounds cost at the same rate?
Check Answer
Unit price:
\[ 18.40\div8=2.30. \]Then
\[ 2.30(12)=27.60. \]Answer:
\[ \boxed{\$27.60}. \]6. Scale Problem
A drawing uses the scale
\[ 2\text{ cm}:5\text{ m}. \]What real length corresponds to 14 cm?
Check Answer
\[ \frac{2}{5}=\frac{14}{x}. \] \[ 2x=70, \] \[ \boxed{x=35\text{ m}}. \]7. Domain Restriction
Solve:
\[ \frac{x+1}{x-2} = \frac{5}{x-2}. \]Check Answer
Restriction:
\[ x\ne2. \]Clearing the denominator gives
\[ x+1=5, \]so
\[ x=4. \]This is allowed, so
\[ \boxed{x=4}. \]Key Ideas
- Proportions can be solved by scaling, unit rates, equivalent fractions, or cross products.
- The simplest valid method is often better than automatically using cross multiplication.
- Cross multiplication is just denominator clearing based on equality properties.
- Corresponding quantities and units must be aligned before solving.
- Variable denominators create domain restrictions that must be recorded before simplification.
- Complex fractions can often be simplified before solving the proportion.
- Reducing ratios first can reveal an easy scale factor.
- Proportional models require a constant multiplicative relationship.
- Not every linear or numerical relationship should be modeled with a proportion.
- Verification by substitution or cross products confirms the final result.
- Estimation provides a useful check against reversed ratios and arithmetic errors.
- Proportion solving connects fraction equivalence, equations, units, scale, percent, and applied modeling.
Where This Leads
Solving proportions turns multiplicative structure into a practical algebraic tool.
The same methods will be used in direct variation, scale and similarity, percent, unit conversion, motion, work rates, mixture problems, financial models, and many situations where a constant relationship lets one known pair determine another.