Find the hidden “equal-quantity” relationship, and the equation writes itself.
Every word problem hides a sentence that says “this amount equals that amount.” A meeting problem hides “the two distances together make the whole road.” An age problem hides “in a few years one age is twice the other.” Once you spot that one equal-quantity relationship, you give the unknown a name, write each side as an expression, set them equal — and the algebra you learned in Lesson 10.2 finishes the job. This lesson is about finding that sentence.
We keep the same color habit all the way through: the unknown is violet, one side or first description is teal, the other side or second description is amber, and the answer is red. When the picture and the equation share colors, the story and the math become the same thing.
The hard part of a word problem is never the algebra — it is the translating. So we follow a fixed routine that does the translating for us. At the center of it sits one idea: somewhere in the story, two different descriptions name the same amount. Find that one equal-quantity relationship and the equation almost writes itself.
Here is the routine, the same five steps every time:
Let’s run the whole routine on a small puzzle: “Three less than twice a number is seventeen. What is the number?”
1. Read for the relationship. Two descriptions of one amount: “three less than twice the number” is “seventeen.” The word is means equals.
2. Name the unknown. Let x = the number.
3. Write both sides. “Twice the number” is 2x; “three less than” it is 2x − 3. The other side is 17. Set them equal: 2x − 3 = 17.
4. Solve, and 5. check — below.
| 2x − 3 = 17 | the equation |
| 2x = 20 | add 3 to both sides |
| x = 10 | divide both sides by 2 |
Check against the story. Twice 10 is 20; three less than 20 is 17. ✓ Answer in words: the number is 10.
“Three less than twice the number” is 2x − 3, not 3 − 2x. The amount you subtract comes second. Read “A less than B” as “B minus A.”
Pick a target value and a “twice-then-subtract” recipe; watch the equation build and the number fall out. The check at the bottom must always say ✓.
Most translating is just a dictionary. English phrases turn into algebra one chunk at a time. Learn the chunks and long sentences stop being scary.
| English | Algebra | English | Algebra |
|---|---|---|---|
| 5 more than x | x + 5 | twice x | 2x |
| 7 less than x | x − 7 | half of x | x2 |
| in 4 years (age) | x + 4 | 4 years ago | x − 4 |
| the sum is 20 | … = 20 | A is twice B | A = 2B |
Age problems are the classic place to use these. The trick: everyone ages at the same rate. If t years pass, every person’s age goes up by t — so you add the same t to both people.
Maria is 24 and her son is 6. In how many years will Maria be twice as old as her son?
Relationship: at the future moment, Maria’s age equals twice the son’s age. Name it: let t = the number of years from now. In t years Maria is 24 + t and the son is 6 + t. The relationship “Maria is twice the son” becomes 24 + t = 2(6 + t).
| 24 + t = 2(6 + t) | the equation |
| 24 + t = 12 + 2t | distribute the 2 |
| 12 = t | subtract 12 and t from both sides |
| t = 12 | read it off |
Check. In 12 years Maria is 36 and her son is 18, and 36 = 2·18. ✓ Answer: in 12 years.
“Maria is twice as old as her son” means the son’s age gets the 2: Maria = 2 · son. Writing 2 · Maria = son makes the older person younger — a sign you doubled the wrong side.
One formula rules every motion problem: distance = speed × time, written d = r·t. Whatever the story, you build each traveler’s distance as speed times time, then look for the equal-quantity relationship between those distances.
Two shapes show up again and again. In a meeting problem the travelers move toward each other, so their distances add up to the gap between them. In a catch-up problem one chases the other, so at the moment of catching their distances are equal.
Two towns are 300 km apart. Cars leave at the same time, driving toward each other at 60 km/h and 40 km/h. When do they meet?
Relationship: together they cover the whole road, so car 1’s distance + car 2’s distance = 300. Name it: let t = hours until they meet. Car 1 goes 60t, car 2 goes 40t.
| 60t + 40t = 300 | distances add to the gap |
| 100t = 300 | combine like terms |
| t = 3 | divide both sides by 100 |
Check. In 3 hours car 1 goes 180 km and car 2 goes 120 km; 180 + 120 = 300. ✓ Answer: they meet after 3 hours.
A walker leaves at 5 km/h. One hour later a cyclist sets off down the same road at 15 km/h. When does the cyclist catch the walker?
Relationship: at the catch they are at the same place, so cyclist’s distance = walker’s distance. Name it: let t = hours the cyclist rides. By then the walker has been going t + 1 hours. So 15t = 5(t + 1).
| 15t = 5(t + 1) | distances are equal |
| 15t = 5t + 5 | distribute the 5 |
| 10t = 5 | subtract 5t from both sides |
| t = 0.5 | divide both sides by 10 |
Check. The cyclist rides 15 · 0.5 = 7.5 km; the walker walks 5 · 1.5 = 7.5 km. Same spot. ✓ Answer: the cyclist catches up after half an hour (the walker is 1.5 hours into the trip).
Set the two speeds and the gap; the road shows where each car is, and the equation solves for the meeting time. Watch the two distances always add to the gap.
How can two pipes filling a tank, or two workers painting a fence, share one job? The trick is to call the whole job 1 — one full tank, one finished fence. Then if a worker finishes alone in a hours, in one hour they finish 1a of the job. That fraction is their rate, and rates of helpers working together simply add.
The equal-quantity relationship is: (rate 1 + rate 2) × time = 1 whole job. This is exactly the “clear the fractions” move from Lesson 10.2.
Pipe A fills a tank in 6 hours; pipe B fills it in 3 hours. Open both — how long to fill the tank?
Rates: A does 16 per hour, B does 13 per hour. Name it: let t = hours to fill it together. The work done is rate × time, and it must equal one whole tank:
| (16 + 13)t = 1 | work done = 1 whole job |
| (16 + 26)t = 1 | common denominator 6 |
| 12·t = 1 | add the rates: 36 = 12 |
| t = 2 | multiply both sides by 2 |
Check. In 2 hours A fills 26 = 13 and B fills 23; 13 + 23 = 1 whole tank. ✓ Answer: 2 hours.
Two helpers are faster than either one alone, so the answer must be less than the smaller time (here, under 3 hours). Adding the times (6 + 3 = 9) or averaging them (4.5) makes them slower — clearly wrong. Add the rates, never the times.
Money and mixtures run on three everyday formulas. Each one is just an equal-quantity relationship dressed in dollars or milliliters.
| Setting | Relationship |
|---|---|
| buying & selling | cost + profit = selling price |
| simple interest | interest = principal × rate × time (I = P·r·t) |
| mixtures | concentration = solute ÷ total solution |
The percent idea is the same in all three: a percent is always a percent of some base amount. A 25%-off discount takes away 25% of the list price; 5% interest earns 5% of the principal; a 30% solution is 30% solute of the whole solution.
A jacket is marked $80 and sold at 25% off. What is the selling price?
The discount is 25% of $80, so selling price = list − discount:
80 − 0.25·80 = 80 − 20 = $60.
A neat shortcut: paying after 25% off means paying 75% of the list, so 0.75 · 80 = $60 too — same answer, fewer steps.
Mixtures hide a beautiful trick: when you add pure water, the amount of salt never changes — only the total grows. So you anchor the equation on the thing that stays fixed.
How much pure water must you add to 200 mL of a 30% salt solution to dilute it to 20%?
Anchor on the salt. The salt amount is fixed at 0.30 · 200 = 60 mL. Name it: let x = mL of water to add; the new total is 200 + x. The new solution is 20% salt, so its salt is 0.20(200 + x). That salt is the same 60 mL:
| 60 = 0.20(200 + x) | salt is unchanged |
| 300 = 200 + x | divide both sides by 0.20 |
| x = 100 | subtract 200 from both sides |
Check. Total 300 mL with 60 mL salt: 60 ÷ 300 = 0.20 = 20%. ✓ Answer: add 100 mL of water.
Slide in pure water. The dark band of salt never changes; only the total grows and the percent drops. Find the water that lands you on the target.
Every word problem hides one equal-quantity relationship — two descriptions of the same amount. Read for it, name the unknown with units, write each side as an expression and set them equal, solve with the moves from Lesson 10.2, then check against the story and answer in words. The relationship just wears different costumes: twice as old (age), distances add or distances equal (motion, from d = r·t), rates add to one whole job (work), and percent of a base (profit, interest, mixtures — where the fixed quantity, like the salt, anchors the equation).
So far one unknown has been enough. But “two numbers add to 10 and differ by 4” really needs two letters. In Lesson 10.4 — Linear Equations in Two Unknowns and Systems a single equation becomes a whole line of solutions, and a pair of conditions becomes the place where two lines cross.
Work each one out first, then open the answer to check your thinking. They run easy → hard.
Six questions to lock it in. Tap the answer you think is right.
This lesson serves the Common Core word-problem standards: 7.EE.B.3 (solve multi-step real-life problems with the four operations), 7.EE.B.4a (construct and solve px + q = r from a word problem), and 7.RP.A.3 (percent problems — markup, discount, and simple interest). The single most common misconception is jumping straight to an equation without finding the equal-quantity relationship — students grab numbers and operations in reading order and write something like 60 + 40 = 300t, or set 2·Maria = son. The antidote is to make Step 1 explicit and verbal every time: before any algebra, say out loud the one sentence “______ equals ______” in plain English, then translate that sentence. Pair it with the habit of always checking the answer back in the story (and a sanity check on size — two workers must beat the faster one; a dilution must lower the percent) so an equation that contradicts common sense gets caught.