Every technique in Linear Equations solves an equation once it’s already written down. The actual difficulty in a word problem happens before that — turning a paragraph of English into an equation in the first place. This lesson gives that translation step a repeatable process, then works through the problem types that show up most often: numbers, geometry, motion, mixtures, and interest.
Every worked example below ends the same way: solve the equation using last lesson’s techniques, then translate the number back into an answer to the actual question — the two steps are equally important, and skipping the second one is one of the most common ways to lose credit on an otherwise correct solution.
A Five-Step Process for Word Problems
- Read the problem and identify what’s being asked. Underline or restate the actual question in your own words before writing anything.
- Assign a variable to the unknown. Let the variable represent the specific quantity the problem asks for, and write every other unknown quantity in terms of that same variable.
- Write an equation that models the relationship described. Translate the sentence describing how the quantities relate into algebra, using the table below.
- Solve the equation using the techniques from Linear Equations.
- Check the solution against the original problem, not just the equation — an answer that solves the equation but doesn’t make sense in context (a negative length, a fractional number of coins) means the equation needs to be fixed, not the arithmetic.
Translating Words into Algebra
| Phrase | Algebra |
|---|---|
| A number increased by 5 | \(x + 5\) |
| 5 less than a number | \(x - 5\) |
| Twice a number | \(2x\) |
| A number decreased by 5, then doubled | \(2(x-5)\) |
| The sum of two consecutive integers | \(x + (x+1)\) |
| Half of a number | \(\dfrac{x}{2}\) |
| 5 more than three times a number | \(3x + 5\) |
| A second number is 5 more than twice the first | second number \(= 2x + 5\) |
The single most useful habit here is writing every unknown quantity in terms of the one variable assigned in step 2 — a problem with three unknowns almost never needs three variables, just three expressions built from one.
Number and Consecutive Integer Problems
Consecutive integers are integers that follow one after another, like \(7, 8, 9\); if the first is \(x\), the next is \(x+1\), and the one after that is \(x+2\). Consecutive even (or odd) integers skip by \(2\) instead of \(1\): if the first is \(x\), the next is \(x+2\), then \(x+4\).
$$ \text{Three consecutive integers: } x, \; x+1, \; x+2 \qquad \text{Three consecutive even integers: } x, \; x+2, \; x+4 $$
Geometry Applications
The most common geometry word problems use the perimeter formula for a rectangle, \(P = 2\ell + 2w\), where one side is described in terms of the other.
$$ \text{“length is 3 more than twice the width”} \;\Longrightarrow\; \ell = 2w + 3 $$
Substituting an expression like that into the perimeter formula turns two unknowns into one, which is exactly what makes the resulting equation linear and solvable.
Distance, Rate, and Time
The relationship \(d = rt\) (distance equals rate times time) drives every motion problem. Most set up either two objects traveling toward each other (their distances add to a known total) or the same distance covered two different ways (the two expressions for distance are set equal).
$$ d = rt \qquad \text{two objects meeting: } d_{1} + d_{2} = \text{total distance} $$
Percent and Mixture Problems
Percent problems convert a percentage to a decimal and multiply it by the base quantity: ”\(20\%\) of \(x\)” becomes \(0.20x\). Mixture problems extend this by adding two such expressions together, one for each ingredient being combined, and setting the total equal to the final mixture’s amount and concentration.
$$ 20\% \text{ of } x = 0.20x \qquad \text{mixture: } (\text{amount}_{1})(\text{concentration}_{1}) + (\text{amount}_{2})(\text{concentration}_{2}) = (\text{total amount})(\text{final concentration}) $$
Simple Interest
$$ I = Prt $$
\(I\) is the interest earned, \(P\) is the principal invested, \(r\) is the annual rate written as a decimal, and \(t\) is time in years. Most simple-interest word problems supply three of these four quantities and ask for the fourth, which turns the formula directly into a linear equation.
Worked Example A: Consecutive Integers
The sum of three consecutive integers is \(72\). Find the integers.
Step 1 — assign a variable. Let the smallest integer be \(x\); the three integers are \(x\), \(x+1\), and \(x+2\).
Step 2 — write an equation for their sum:
$$ x + (x+1) + (x+2) = 72 $$
Step 3 — solve:
$$ 3x + 3 = 72 \;\Longrightarrow\; 3x = 69 \;\Longrightarrow\; x = 23 $$
Step 4 — answer the actual question. The three integers are \(23\), \(24\), and \(25\).
Step 5 — check: \(23+24+25 = 72\). ✓
Worked Example B: Geometry Perimeter
A rectangle’s length is \(5\) feet more than twice its width, and its perimeter is \(46\) feet. Find the width and the length.
Step 1 — assign a variable. Let the width be \(w\); the length is \(2w+5\).
Step 2 — substitute into the perimeter formula:
$$ P = 2\ell + 2w \;\Longrightarrow\; 46 = 2(2w+5) + 2w $$
Step 3 — solve:
$$ 46 = 4w + 10 + 2w \;\Longrightarrow\; 46 = 6w + 10 \;\Longrightarrow\; 36 = 6w \;\Longrightarrow\; w = 6 $$
Step 4 — find the length and check it makes sense. The length is \(2(6)+5 = 17\) feet. Both dimensions are positive, which they must be.
Step 5 — check: \(2(17) + 2(6) = 34 + 12 = 46\). ✓
Worked Example C: Distance, Rate, and Time
Two cyclists start from the same point and ride in opposite directions, one at \(12\) mph and the other at \(15\) mph. How long until they are \(54\) miles apart?
Step 1 — assign a variable. Let \(t\) be the time in hours until they are \(54\) miles apart.
Step 2 — write an expression for each distance and add them:
$$ 12t + 15t = 54 $$
Step 3 — solve:
$$ 27t = 54 \;\Longrightarrow\; t = 2 $$
Step 4 — check the answer makes sense in context. \(2\) hours is a reasonable amount of time, and the units work out (mph \(\times\) hours \(=\) miles).
Step 5 — check: \(12(2) + 15(2) = 24 + 30 = 54\). ✓
Worked Example D: A Percent Mixture Problem
How many liters of a \(40\%\) acid solution must be mixed with \(10\) liters of a \(10\%\) acid solution to make a \(25\%\) acid solution?
Step 1 — assign a variable. Let \(x\) be the number of liters of the \(40\%\) solution.
Step 2 — write an equation balancing total acid before and after mixing:
$$ 0.40x + 0.10(10) = 0.25(x + 10) $$
Step 3 — solve:
$$ 0.40x + 1 = 0.25x + 2.5 \;\Longrightarrow\; 0.15x = 1.5 \;\Longrightarrow\; x = 10 $$
Step 4 — answer the question. \(10\) liters of the \(40\%\) solution are needed.
Step 5 — check: \(0.40(10) + 0.10(10) = 4 + 1 = 5\) liters of acid, and \(0.25(10+10) = 0.25(20) = 5\) liters of acid. ✓
Common Mistakes to Avoid
- Solving for \(x\) and stopping there. If the problem asks for the length, the width, or an integer other than the first one, \(x\) alone usually isn’t the final answer — always translate back to the quantity the problem actually asked for.
- Using a different variable for every unknown. This usually produces one equation with two unknowns, which can’t be solved on its own. Write every unknown in terms of the same variable instead.
- Skipping the units. \(d=rt\) only comes out right when rate and time use matching units — mixing hours with minutes, or miles with feet, gives a numerically wrong answer even with perfectly correct algebra.
- Forgetting to convert a percent to a decimal. \(20\%\) is \(0.20\) in an equation, not \(20\).
- Accepting a solution that doesn’t make sense in context. A negative length, a fractional number of people, or a physically impossible rate means the equation was set up wrong — go back to step 3, not step 4.
- Setting up the perimeter formula with both sides as the same variable. A rectangle’s length and width are generally different; conflating them turns a rectangle problem into a square problem by accident.
- Adding rates instead of adding distances. In the meeting-cyclists example, the two distances add to the total, not the two rates by themselves — rate needs to be multiplied by time first.
Where This Shows Up Later
- Systems of equations. Word problems with two genuinely independent unknowns, rather than one expressed in terms of the other, are usually solved as a system instead of forcing everything into one variable.
- Quadratic applications. Area problems, projectile motion, and some geometry problems use this exact five-step process but end in a quadratic equation instead of a linear one.
- Equations with More Than One Variable. Formulas like \(d=rt\), \(I=Prt\), and \(P=2\ell+2w\) all reappear there, solved for a different letter instead of a number.
- Rational equation applications. Work-rate problems (“how long would it take two people working together”) use this same five-step setup, applied to a rational equation instead of a linear one.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. The sum of two consecutive integers is \(45\). Find the integers.
Show answer
Step 1 — assign a variable. Let the smaller integer be \(x\); the next is \(x+1\).
Step 2 — write an equation:
$$ x + (x+1) = 45 $$
Step 3 — solve:
$$ 2x + 1 = 45 \;\Longrightarrow\; 2x = 44 \;\Longrightarrow\; x = 22 $$
Answer: the integers are \(22\) and \(23\).
Problem 2. Four times a number, decreased by \(7\), is \(29\). Find the number.
Show answer
Step 1 — assign a variable and write an equation:
$$ 4x - 7 = 29 $$
Step 2 — solve:
$$ 4x = 36 \;\Longrightarrow\; x = 9 $$
Answer: the number is \(9\).
Problem 3. A rectangle’s length is \(4\) cm more than its width, and its perimeter is \(32\) cm. Find the width and the length.
Show answer
Step 1 — assign a variable. Let the width be \(w\); the length is \(w+4\).
Step 2 — substitute into the perimeter formula:
$$ 2(w+4) + 2w = 32 $$
Step 3 — solve:
$$ 2w + 8 + 2w = 32 \;\Longrightarrow\; 4w + 8 = 32 \;\Longrightarrow\; 4w = 24 \;\Longrightarrow\; w = 6 $$
Answer: the width is \(6\) cm and the length is \(6+4 = 10\) cm.
Problem 4. The sum of three consecutive even integers is \(96\). Find the integers.
Show answer
Step 1 — assign a variable. Let the smallest even integer be \(x\); the next two are \(x+2\) and \(x+4\).
Step 2 — write an equation:
$$ x + (x+2) + (x+4) = 96 $$
Step 3 — solve:
$$ 3x + 6 = 96 \;\Longrightarrow\; 3x = 90 \;\Longrightarrow\; x = 30 $$
Answer: the integers are \(30\), \(32\), and \(34\).
Problem 5. Two trains leave the same station at the same time, traveling in opposite directions, one at \(50\) mph and the other at \(70\) mph. After how many hours are they \(360\) miles apart?
Show answer
Step 1 — assign a variable. Let \(t\) be the number of hours.
Step 2 — write an equation adding the two distances:
$$ 50t + 70t = 360 $$
Step 3 — solve:
$$ 120t = 360 \;\Longrightarrow\; t = 3 $$
Answer: \(3\) hours.
Problem 6. \(\$8{,}000\) is invested at a simple annual interest rate. After \(2\) years, the interest earned is \(\$960\). Find the interest rate.
Show answer
Step 1 — substitute into the simple interest formula:
$$ I = Prt \;\Longrightarrow\; 960 = 8000 \cdot r \cdot 2 $$
Step 2 — solve:
$$ 960 = 16000r \;\Longrightarrow\; r = 0.06 $$
Answer: the interest rate is \(0.06\), or \(6\%\).
Problem 7. How many liters of a \(60\%\) alcohol solution must be added to \(5\) liters of a \(20\%\) alcohol solution to make a \(40\%\) solution?
Show answer
Step 1 — assign a variable. Let \(x\) be the number of liters of the \(60\%\) solution.
Step 2 — write an equation balancing total alcohol:
$$ 0.60x + 0.20(5) = 0.40(x+5) $$
Step 3 — solve:
$$ 0.60x + 1 = 0.40x + 2 \;\Longrightarrow\; 0.20x = 1 \;\Longrightarrow\; x = 5 $$
Answer: \(5\) liters.
Problem 8. A number increased by \(12\) equals three times the number. Find the number.
Show answer
Step 1 — write an equation:
$$ x + 12 = 3x $$
Step 2 — solve:
$$ 12 = 2x \;\Longrightarrow\; x = 6 $$
Answer: the number is \(6\).
Problem 9. Twelve less than five times a number is the same as the number increased by \(20\). Find the number.
Show answer
Step 1 — write an equation:
$$ 5x - 12 = x + 20 $$
Step 2 — solve:
$$ 4x = 32 \;\Longrightarrow\; x = 8 $$
Answer: the number is \(8\).
Problem 10. The perimeter of a triangle is \(39\) cm. The second side is twice the first, and the third side is \(3\) cm more than the first. Find all three side lengths.
Show answer
Step 1 — assign a variable. Let the first side be \(x\); the second is \(2x\) and the third is \(x+3\).
Step 2 — write an equation for the perimeter (sum of the three sides):
$$ x + 2x + (x+3) = 39 $$
Step 3 — solve:
$$ 4x + 3 = 39 \;\Longrightarrow\; 4x = 36 \;\Longrightarrow\; x = 9 $$
Answer: the sides are \(9\) cm, \(18\) cm, and \(12\) cm. Check: \(9+18+12=39\). ✓
Quick Reference
| Word problem type | Core relationship |
|---|---|
| Consecutive integers | \(x,\; x+1,\; x+2,\; \dots\) |
| Consecutive even or odd integers | \(x,\; x+2,\; x+4,\; \dots\) |
| Rectangle perimeter | \(P = 2\ell + 2w\) |
| Distance, rate, time | \(d = rt\) |
| Percent of a quantity | percent (as a decimal) times the quantity |
| Mixture | (amount)(concentration), added across ingredients, equals (total amount)(final concentration) |
| Simple interest | \(I = Prt\) |
| Every problem | Assign one variable, write every other unknown in terms of it, solve, then check against the original question |
Once a word problem’s five steps feel automatic, the equations themselves lean directly on Linear Equations and, for formulas solved for a different letter, Equations with More Than One Variable. Revisit Solutions and Solution Sets for the vocabulary behind checking a candidate answer, or browse the rest of the Algebra lessons as new ones publish.