Mathematics II · Calculator allowed

Writing and Solving Linear Equations from Word Problems

The hardest part of a word problem usually is not the math, it is turning the words into an equation. Here is a simple, repeatable way to do exactly that.

On the calculator part of the CAEC, a lot of questions are dressed up as little stories: a phone plan, a delivery fee, splitting a bill, a savings goal. Your calculator will happily do the arithmetic, but first you have to translate the story into an equation it can chew on.

Good news: there is a reliable, three-part process that works on almost every one of these. Once you have it, word problems stop feeling like mind-reading and start feeling like a checklist. Let's walk through it together.

The process: translate, solve, check

Every word-problem-into-equation question follows the same three moves. Do them in order and you will rarely get lost:

1. Translate. Pick a letter for the unknown (the thing you are asked to find), then rewrite the sentence as an equation, phrase by phrase.
2. Solve. Use inverse operations to get the variable by itself. Whatever you do to one side, do to the other.
3. Check. Put your answer back into the original equation (or the original sentence) and make sure both sides match.

The key habit: always start by writing "Let x = ..." in plain words. If you are crystal clear about what your variable stands for, the rest of the equation almost writes itself.

A mini phrasebook: words to math

English has a lot of ways to say the same operation. These are the ones that come up again and again in word problems:

These words...	...mean this
sum, total, more than, increased by, plus	+
difference, less than, decreased by, fewer, minus	−
product, times, of, twice, per, each	×
quotient, divided by, split equally, per	÷
is, was, equals, gives, results in, the same as	=

Watch the word order with "less than." "5 less than a number" is x − 5, not 5 − x. The thing after "than" comes first.

Worked example #1: a one-step equation

A pair of running shoes is on sale for $68. That is $22 off the regular price. What was the regular price?

Translate: Let p = the regular price. The sale price is the regular price minus $22, and that equals $68: p − 22 = 68.
Solve: The opposite of subtracting 22 is adding 22. Add 22 to both sides.
Check: 90 − 22 = 68. That matches the sale price, so we are good.

  p − 22 = 68
  p − 22 + 22 = 68 + 22   (add 22 to both sides)
  p = 90

Answer: the regular price was $90. Notice we undid the subtraction with its opposite, addition.

Worked example #2: a two-step equation

A plumber charges a $60 call-out fee plus $45 for each hour of work. The total bill was $240. How many hours did the plumber work?

Translate: Let h = the number of hours. The bill is the flat $60 plus $45 per hour, and the total is $240: 45h + 60 = 240.
Solve, step 1: Subtract the 60 first (undo the addition), leaving 45h = 180.
Solve, step 2: Divide both sides by 45 (undo the multiplication): h = 4.
Check: 45 × 4 + 60 = 180 + 60 = 240. Matches the bill exactly.

  45h + 60 = 240
  45h + 60 − 60 = 240 − 60   (subtract 60 from both sides)
  45h = 180
  45h ÷ 45 = 180 ÷ 45        (divide both sides by 45)
  h = 4

Answer: the plumber worked 4 hours. With two-step equations, undo the adding/subtracting first, then the multiplying/dividing, the reverse of the order of operations.

Worked example #3: variables on both sides

Gym A charges $25 to join plus $20 a month. Gym B has no joining fee but charges $25 a month. After how many months do the two gyms cost the same?

Translate: Let m = the number of months. Gym A costs 20m + 25; Gym B costs 25m. Set them equal: 20m + 25 = 25m.
Gather the variables: Subtract 20m from both sides so the variable lives on one side only, leaving 25 = 5m.
Solve: Divide both sides by 5: m = 5.
Check: Gym A: 20 × 5 + 25 = 125. Gym B: 25 × 5 = 125. Equal, so 5 months is right.

  20m + 25 = 25m
  20m + 25 − 20m = 25m − 20m   (subtract 20m from both sides)
  25 = 5m
  25 ÷ 5 = 5m ÷ 5              (divide both sides by 5)
  5 = m

Answer: the gyms cost the same after 5 months. When the variable appears on both sides, move all the variable terms to one side first, then finish like a normal two-step equation.

Worked example #4: when one quantity depends on another

Two friends share 84 raffle tickets. Maya has 3 times as many as Leo. How many tickets does each person have?

Translate: Let the smaller share be the variable. Let x = Leo's tickets. Then Maya has 3x. Together they make 84: x + 3x = 84.
Combine like terms: x + 3x = 4x, so 4x = 84.
Solve: Divide both sides by 4: x = 21.
Answer both parts: Leo has 21; Maya has 3 × 21 = 63. Check: 21 + 63 = 84.

  x + 3x = 84
  4x = 84            (combine like terms)
  4x ÷ 4 = 84 ÷ 4    (divide both sides by 4)
  x = 21

Answer: Leo has 21 tickets and Maya has 63. When one amount is described in terms of another, name the variable for the simpler one and build the rest around it. And reread the question, it asked for both amounts, not just x.

Tips that make word problems easier

Write "Let x = ..." every time. Naming the unknown in words is the single biggest thing that keeps you from getting tangled up.
Underline the question. Make sure you know exactly what is being asked before you solve, sometimes the answer is 3x, not x.
Look for the "is." The word "is" (or "was," "equals," "total") is usually where your equals sign goes. Everything before it is one side; everything after is the other.
Undo in reverse. To get the variable alone, peel away the add/subtract first, then the multiply/divide. It is the order of operations run backwards.
Always check by substituting. Put your answer back into the original equation. Your calculator makes this quick, and it catches almost every slip.

Your turn: practice problems

For each one, write "Let x = ...," build the equation, solve it, and check. A calculator is fine. No peeking until you have tried.

A number increased by 17 is 50. What is the number?
A taxi charges a $4 base fare plus $2 per kilometre. The ride cost $30. How many kilometres was the ride?
A jacket costs $12 more than a shirt. Together they cost $76. How much does each item cost?
Plan A costs $40 plus $0.10 per text. Plan B costs $25 plus $0.25 per text. After how many texts do the plans cost the same?

Tap to reveal the answers

1. Let x = the number. x + 17 = 50. Subtract 17: x = 33. Check: 33 + 17 = 50.
2. Let x = kilometres. 2x + 4 = 30. Subtract 4: 2x = 26. Divide by 2: x = 13 km. Check: 2 × 13 + 4 = 30.
3. Let x = the shirt's price; the jacket is x + 12. Then x + (x + 12) = 76, so 2x + 12 = 76. Subtract 12: 2x = 64. Divide by 2: x = 32. The shirt is $32 and the jacket is $44. Check: 32 + 44 = 76.
4. Let x = number of texts. 0.10x + 40 = 0.25x + 25. Subtract 0.10x: 40 = 0.15x + 25. Subtract 25: 15 = 0.15x. Divide by 0.15: x = 100 texts. Check: Plan A = 0.10 × 100 + 40 = $50; Plan B = 0.25 × 100 + 25 = $50.

Why this matters for the CAEC

The calculator part of the CAEC math test is full of applied, real-world problems, phone plans, fees, splitting costs, savings goals. Almost all of them come down to writing one linear equation and solving it. The translate-solve-check habit turns these from intimidating stories into quick, confident points.

Want more practice like this? Our CAEC math guide and the CAEC Ready Workbook are packed with worked examples and practice questions, or start with a free math sample to test yourself.

Disclaimer

This article is a general math tutorial for study purposes. CAEC Ready is an independent study resource and is not affiliated with or endorsed by any government, ministry of education, or official CAEC testing provider.