Linear Equation Word Problems

Word problems are where algebra meets real life. The hardest part is not the solving — it is the translation from English into an equation. Once you have the equation written down, you already know how to solve it. This page gives you a repeatable system for turning any word problem into a linear equation, then walks through the five most common problem types with fully worked examples.

The 5-Step Translation System

Use this framework for every word problem:

1. Read the entire problem. Do not start writing until you understand what is being asked.
2. Define your variable. Write “Let $x$ = …” with a clear description. Every other unknown quantity should be expressed in terms of $x$.
3. Translate the words into an equation. Look for key phrases (see the table below).
4. Solve the equation.
5. Answer the question and check. Substitute back into the original problem’s words (not just the equation) to verify the answer makes sense.

Key Translation Phrases

Important: “5 less than a number” is $x - 5$, not $5 - x$. The phrase “less than” reverses the order.

Type 1: Number Problems

These are pure translation exercises — great for building confidence.

Example 1: Consecutive Integers

The sum of three consecutive integers is 72. Find the integers.

Step 1 — Define the variable:

Let $x$ = the first integer. Then the next two are $x + 1$ and $x + 2$.

Step 2 — Write the equation:

$x + (x + 1) + (x + 2) = 72$

Step 3 — Solve:

$3x + 3 = 72$

$3x = 69$

$x = 23$

Step 4 — Answer and check:

The integers are 23, 24, and 25. Check: $23 + 24 + 25 = 72$ . Correct.

Example 2: A Number Relationship

One number is 4 more than three times another number. Their sum is 36. Find the numbers.

Let $x$ = the smaller number. The larger number is $3x + 4$.

$x + (3x + 4) = 36$

$4x + 4 = 36$

$4x = 32$

$x = 8$

The larger number: $3(8) + 4 = 28$.

Answer: The two numbers are 8 and 28. Check: $8 + 28 = 36$ and $28 = 3(8) + 4$ . Correct.

Type 2: Age Problems

Age problems involve relationships between people’s ages at different points in time.

Example 3: Present Ages

Maria is 5 years older than twice her daughter’s age. The sum of their ages is 44. How old are they?

Let $x$ = the daughter’s age. Maria’s age is $2x + 5$.

$x + (2x + 5) = 44$

$3x + 5 = 44$

$3x = 39$

$x = 13$

Maria’s age: $2(13) + 5 = 31$.

Answer: The daughter is 13 and Maria is 31. Check: $13 + 31 = 44$ and $31 = 2(13) + 5$ . Correct.

Example 4: Future Ages

A father is currently 40 years old and his son is 12. In how many years will the father be exactly twice as old as his son?

Let $x$ = the number of years from now.

In $x$ years: father is $40 + x$, son is $12 + x$.

$40 + x = 2(12 + x)$

$40 + x = 24 + 2x$

Subtract $x$: $40 = 24 + x$

Subtract 24: $x = 16$

Answer: In 16 years, the father will be 56 and the son will be 28. Check: $56 = 2(28)$ . Correct.

Type 3: Coin and Money Problems

These problems involve different denominations or prices adding up to a total.

Example 5: Mixed Coins

A jar contains only quarters and dimes. There are 30 coins worth a total of $5.70. How many of each type?

Let $x$ = the number of quarters. Then $30 - x$ = the number of dimes.

Value equation (in cents to avoid decimals):

$25x + 10(30 - x) = 570$

$25x + 300 - 10x = 570$

$15x + 300 = 570$

$15x = 270$

$x = 18$

Dimes: $30 - 18 = 12$.

Answer: There are 18 quarters and 12 dimes. Check: $18(0.25) + 12(0.10) = 4.50 + 1.20 = 5.70$ . Correct.

Example 6: Ticket Sales

An event sells adult tickets for $12 and student tickets for $7. A total of 200 tickets were sold for $1,950. How many of each type were sold?

Let $x$ = adult tickets. Student tickets = $200 - x$.

$12x + 7(200 - x) = 1950$

$12x + 1400 - 7x = 1950$

$5x + 1400 = 1950$

$5x = 550$

$x = 110$

Student tickets: $200 - 110 = 90$.

Answer: 110 adult and 90 student tickets. Check: $110(12) + 90(7) = 1320 + 630 = 1950$ . Correct.

Type 4: Perimeter and Geometry Problems

These use known geometric relationships to set up equations.

Example 7: Rectangle Perimeter

The length of a rectangle is 3 cm more than twice the width. The perimeter is 54 cm. Find the dimensions.

Let $w$ = the width. The length is $2w + 3$.

Perimeter formula: $P = 2l + 2w$

$2(2w + 3) + 2w = 54$

$4w + 6 + 2w = 54$

$6w + 6 = 54$

$6w = 48$

$w = 8$

Length: $2(8) + 3 = 19$.

Answer: Width = 8 cm, Length = 19 cm. Check: $2(19) + 2(8) = 38 + 16 = 54$ . Correct.

Example 8: Triangle Angles

The second angle of a triangle is twice the first angle. The third angle is 15 degrees more than the first. Find all three angles.

Let $x$ = the first angle. Second angle = $2x$. Third angle = $x + 15$.

The sum of angles in a triangle is $180°$:

$x + 2x + (x + 15) = 180$

$4x + 15 = 180$

$4x = 165$

$x = 41.25$

Answer: The angles are 41.25°, 82.5°, and 56.25°. Check: $41.25 + 82.5 + 56.25 = 180$ . Correct.

Type 5: Percent Mixture Problems

These involve combining two quantities with different concentrations or values.

Example 9: Mixing Solutions

A chemist needs 60 mL of a 40% acid solution. She has 30% and 50% solutions available. How much of each should she mix?

Let $x$ = mL of 30% solution. Then $60 - x$ = mL of 50% solution.

The acid in the mixture must equal 40% of 60 mL:

$0.30x + 0.50(60 - x) = 0.40(60)$

$0.30x + 30 - 0.50x = 24$

$-0.20x + 30 = 24$

$-0.20x = -6$

$x = 30$

Answer: Mix 30 mL of the 30% solution with 30 mL of the 50% solution. Check: $0.30(30) + 0.50(30) = 9 + 15 = 24$ mL of acid, and $\frac{24}{60} = 0.40$, or 40% . Correct.

Real-World Application: Retail — Pricing and Discount

A store manager marks up items by 40% over wholesale cost. During a sale, the marked-up price is discounted by 15%. If the sale price of an item is $47.60, what was the wholesale cost?

Let $x$ = wholesale cost.

Marked-up price = $1.40x$ (a 40% markup).

Sale price = $1.40x \times 0.85$ (a 15% discount applied to the markup).

$1.40x \times 0.85 = 47.60$

$1.19x = 47.60$

$x = \frac{47.60}{1.19} = 40$

Answer: The wholesale cost was $40.00. Check: markup gives $40 \times 1.40 = 56$, then the discount gives $56 \times 0.85 = 47.60$ . Correct.

Real-World Application: Carpentry — Material Estimation

A carpenter is building a rectangular deck. The total perimeter of railing needed is 56 feet. The length must be 4 feet more than the width to fit the house layout. The railing costs $18 per linear foot. What are the deck dimensions and the total railing cost?

Let $w$ = width in feet. Length = $w + 4$.

$2(w + 4) + 2w = 56$

$2w + 8 + 2w = 56$

$4w + 8 = 56$

$4w = 48$

$w = 12$

Length: $12 + 4 = 16$ feet.

Railing cost: $56 \times 18 = 1{,}008$

Answer: Width = 12 ft, Length = 16 ft, Total railing cost = $1,008. Check: $2(16) + 2(12) = 32 + 24 = 56$ feet . Correct.

Real-World Application: Nursing — IV Dosage Timing

A nurse needs to administer 750 mL of saline. The drip rate for the first part of the infusion is 150 mL/hr. After the first 300 mL, the doctor orders the rate reduced to 100 mL/hr for the remainder. What is the total infusion time?

Time for first 300 mL at 150 mL/hr:

$t_1 = \frac{300}{150} = 2 \text{ hours}$

Remaining volume: $750 - 300 = 450$ mL at 100 mL/hr:

$t_2 = \frac{450}{100} = 4.5 \text{ hours}$

Total time:

$t = t_1 + t_2 = 2 + 4.5 = 6.5 \text{ hours}$

Answer: The total infusion time is 6.5 hours (6 hours and 30 minutes). The nurse documents the expected completion time for the care team.

Common Mistakes to Avoid

1. Defining the variable too vaguely. “Let $x$ = the number” is not enough. Write “Let $x$ = the width in cm” so you know exactly what $x$ represents.
2. Reversing “less than.” “5 less than $x$” is $x - 5$, not $5 - x$. Always ask: “Which quantity is being reduced?”
3. Forgetting that all unknowns must be expressed in terms of one variable. If there are two unknowns and one equation, the second unknown must be written in terms of the first (e.g., “the other number is $30 - x$”).
4. Mixing units. If you express coin values in cents in one term and dollars in another, the equation will be wrong. Choose one unit and stick with it.
5. Not answering the actual question. If the problem asks “How many dimes?”, do not stop after finding $x$ (the number of quarters). Compute $30 - x$ and state the answer clearly.
6. Skipping the check. Substitute your answer into the original words, not just the equation. This catches translation errors that the equation check would miss.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The hardest part of word problems is translation, not solving — use the 5-step system every time
• Define your variable precisely with units and a clear description
• Express all unknowns in terms of one variable before writing the equation
• Watch for “less than” — it reverses the order ($x - 5$, not $5 - x$)
• Use cents instead of dollars for coin problems to avoid decimals
• For mixture problems, the equation comes from: amount of substance in part 1 + amount in part 2 = amount in the mixture
• Always check your answer against the original words of the problem, not just the equation

Return to Algebra for more topics in this section.