When two things are unknown, name two letters — and the idea scales to three.
Some stories carry two mysteries at once. Twenty tickets were sold for a total of $139 — but how many were adult tickets, and how many were children's? You could wrestle the whole thing onto one letter, but it is far smoother to give each mystery its own name: let a be the adults and c be the children. Two unknowns, two facts, two equations — a system, exactly the machinery you built in Lessons 10.4 and 10.5. In this lesson you turn that machinery on real problems, and then you do something new: you add a third unknown.
The same color habit guides us the whole way down. The unknowns are violet, the first equation is teal, the second equation is amber, and the final solution is red. When a third unknown joins, you will watch three equations fold into two, then two into one — and the answer drops out.
Back in Lesson 10.3 you met the five-step frame for word problems: read, name, translate, solve, check. Nothing about that frame changes here. The only difference is what happens at the name step. When a story has two unknown quantities tangled together by two equal-quantity relationships, forcing everything onto a single letter is like tying one shoe with the other foot. Give each unknown its own letter, write the two facts as two equations, and let the system do the work.
How do you spot a two-unknown story? Listen for two separate "totals" or "amounts" you do not know, joined by two sentences that each say this equals that. Tickets are the classic case: you don't know the number of adults or the number of children, and you're told two things — a total count and a total cost.
The story. Adult tickets cost $8, children's tickets cost $5. A booth sold 20 tickets for a total of $139. How many of each were sold?
Name. Let a = number of adult tickets and c = number of children's tickets.
Translate. The count gives a + c = 20. The money gives 8a + 5c = 139 — eight dollars per adult, five per child.
Either method from Lesson 10.5 works. Substitution is tidy here because the count equation rearranges to a = 20 − c with no fractions. Put that into the money equation:
| 8a + 5c = 139 | the cost equation |
| 8(20 − c) + 5c = 139 | replace a with 20 − c |
| 160 − 8c + 5c = 139 | distribute the 8 |
| 160 − 3c = 139 | combine the c-terms |
| −3c = −21 | subtract 160 from both sides |
| c = 7 | divide both sides by −3 |
Then back-substitute: a = 20 − 7 = 13. So 13 adult tickets and 7 children's tickets.
Check — and always check against the original story, not your own equations. Count: 13 + 7 = 20 ✓. Cost: 8·13 + 5·7 = 104 + 35 = 139 ✓. Both facts hold, so the answer is solid.
Two unknowns deserve two letters. The five-step frame is unchanged — you simply translate two relationships into two equations and solve the system.
Set the two prices and the totals. The widget writes the system and solves it — but only whole, non-negative ticket counts make sense, so watch when an answer turns red.
A whole family of puzzles is built from two numbers and two relationships between them. "Their sum is 30 and their difference is 8." "One is three times the other, and together they make 48." Each clue becomes an equation; name the two numbers and you are back in familiar territory.
The richest of these are digit problems, and they hide a trap worth naming out loud. A two-digit number is not its two digits stuck together — it is built by place value. The number whose tens digit is t and whose units digit is u equals 10t + u. Reverse the digits and you get 10u + t. Get that representation right and the algebra is easy; get it wrong and nothing will check.
The story. The two digits of a two-digit number add up to 12. Reversing the digits makes a number that is 18 more than the original. What is the number?
Name. Let t = the tens digit and u = the units digit. The original number is 10t + u.
Translate. Digits sum to 12: t + u = 12. Reversed is 18 more: 10u + t = (10t + u) + 18.
The second equation looks bulky, but it simplifies beautifully. Bring every term to one side:
| 10u + t = 10t + u + 18 | the "reversed is 18 more" fact |
| 10u − u + t − 10t = 18 | move all the unknowns left |
| 9u − 9t = 18 | combine like terms |
| u − t = 2 | divide both sides by 9 |
Now the system is wonderfully simple: t + u = 12 and u − t = 2. Add the two equations and the t cancels: 2u = 14, so u = 7. Then t = 12 − 7 = 5.
The number is 10·5 + 7 = 57. Check: 5 + 7 = 12 ✓; reversed is 75, and 75 − 57 = 18 ✓.
The answer to a digit problem is the number, 57 — not "t = 5." A common slip is to write the value as "57" while accidentally computing 10u + t or just t + u. Always rebuild the number with place value: 10·(tens) + (units).
Slide the sum and the difference of two numbers. The widget shows how adding the two equations finds the larger number and subtracting finds the smaller — the heart of every "sum & difference" puzzle.
Pairs of unknowns turn up everywhere once you look. Two travelers approach each other and you want both speeds. A workshop matches bolts to nuts and you want the count of each. A prize is split by a ratio and you want each share. In every case, two quantities are unknown and two facts pin them down.
The crispest example is a boat fighting a current. Going downstream, the current pushes you along, so your ground speed is boat speed + water speed. Going upstream, the current holds you back, so your ground speed is boat speed − water speed. That single physical idea hands you two equations.
The story. A boat travels 30 km downstream in 2 hours, and the same 30 km upstream in 3 hours. Find the boat's speed in still water and the speed of the current.
Name. Let b = the boat's speed in still water (km/h) and w = the water's speed (km/h).
Translate. Downstream ground speed is distance ÷ time = 30 ÷ 2 = 15, so b + w = 15. Upstream ground speed is 30 ÷ 3 = 10, so b − w = 10.
These two equations are made for elimination. Add them and w vanishes:
| b + w = 15 | downstream |
| b − w = 10 | upstream |
| 2b = 25 | add the two equations (the w's cancel) |
| b = 12.5 | divide both sides by 2 |
Back-substitute into the downstream equation: 12.5 + w = 15, so w = 2.5. The boat moves at 12.5 km/h in still water, and the current runs at 2.5 km/h.
Check: 12.5 + 2.5 = 15 ✓ (downstream), and 12.5 − 2.5 = 10 ✓ (upstream). It is fine — and common — for these answers to be non-whole.
Whenever something helps one way and hinders the other — a current, a tailwind, a moving walkway — name the still speed and the helper speed, then add for one and subtract for the other. Adding the equations cancels the helper instantly.
Choose the downstream and upstream ground speeds. Watch the system solve by adding (to find b) and subtracting (to find w). Set them to 15 and 10 to recover the worked example.
Here is the new idea, and it is a small leap. If one unknown needs one equation, and two unknowns need two, then three unknowns need three conditions. We bundle the three equations together with a single tall brace and call the whole package a linear system in three unknowns.
What is a solution now? With one unknown, a solution was a number. With two, it was an ordered pair (x, y) — a point where two lines crossed. With three unknowns, a solution is an ordered triple (x, y, z), and to count as a solution it must make all three equations true at once. Miss even one and the triple is wrong.
Count your letters, count your equations — you need as many independent conditions as unknowns. Three unknowns ⇒ three equations ⇒ an answer of the form (x, y, z), checked against all three.
Geometrically, each equation like x + 2y + 3z = 14 is now a flat plane floating in three-dimensional space, and the solution is the single point where all three planes meet. You do not need to picture that to solve it — but it explains why three conditions are exactly enough: two planes meet in a line, and the third plane cuts that line at one point.
The strategy is the one you already trust, used twice. Step one: pair the equations up and eliminate one chosen letter, turning a three-unknown system into a two-unknown system. Step two: solve that two-unknown system the ordinary way. Step three: back-substitute up the ladder to recover the letter you eliminated first. Three unknowns become two, two become one — then you climb back up.
Let's solve a full example. Notice that all three equations start with a lone x — that makes x the obvious letter to eliminate first, because subtracting one equation from another cancels it cleanly.
| x + y + z = 6 | equation ① (call it e1) |
| x + 2y + 3z = 14 | equation ② (e2) |
| x + 4y + 9z = 36 | equation ③ (e3) |
Step 1 — knock out x. Subtract e1 from e2, and e2 from e3. Each lone x cancels:
| (e2 − e1): y + 2z = 8 | (2−1)y + (3−1)z = 14−6 |
| (e3 − e2): 2y + 6z = 22 | (4−2)y + (9−3)z = 36−14 |
We now have a clean two-unknown system in y and z. Tidy the second one by dividing by 2: 2y + 6z = 22 becomes y + 3z = 11.
Step 2 — solve the two-unknown system. Subtract the equations to eliminate y:
| y + 3z = 11 | from (e3 − e2), divided by 2 |
| y + 2z = 8 | from (e2 − e1) |
| z = 3 | subtract: (3z − 2z) = (11 − 8) |
Step 3 — climb back up. With z = 3, the equation y + 2z = 8 gives y = 8 − 2·3 = 2. Then e1, x + y + z = 6, gives x = 6 − 2 − 3 = 1. The solution is the triple (1, 2, 3).
Check against all three originals. e1: 1 + 2 + 3 = 6 ✓. e2: 1 + 2·2 + 3·3 = 1 + 4 + 9 = 14 ✓. e3: 1 + 4·2 + 9·3 = 1 + 8 + 27 = 36 ✓. All three hold, so (1, 2, 3) is the answer.
When you make the second pairing, reuse a fresh equation, not the one you've already spent. Here e1 was paired with e2, and e2 with e3 — every original got used. If you pair e1 with e2 twice, you learn the same thing twice and the system won't close.
Press the steps in order to descend the ladder: first cancel x with e2−e1 and e3−e2, then find z, then y, then x. Each rung reveals the next.
Dial in a candidate triple (x, y, z) and the widget substitutes it into all three equations of the worked system. A triple is the solution only when every row turns green. Find (1, 2, 3).
When a story hides two unknowns and offers two facts, name two letters and write two equations — tickets, reversed digits, a boat against the current all fall to the same five-step frame, finished by substitution or elimination and verified against the original story. Add a third unknown and you need a third condition; bundle the three equations in one brace, and a solution becomes an ordered triple (x, y, z). To solve it, eliminate one letter to drop from three unknowns to two, solve that pair, then climb back up by back-substitution — and always check the triple against all three originals.
You have now solved linear equations one letter, two letters, and three letters at a time. In Lesson 10.7 we look back over the whole linear journey and push open the next door: what happens when the unknown is multiplied by itself — the quadratic.
Work each one out first, then open the answer to check your thinking.
Six questions to lock it in. Tap the answer you think is right.
This lesson serves Common Core 8.EE.C.8c (solve real-world and mathematical problems leading to two linear equations in two variables) and reaches toward A.REI.C.6 (solve systems of linear equations, including the high-school extension to three variables, by elimination). The earlier word-problem standard 7.EE.B.4 is also reinforced, now with two unknowns instead of one.
The #1 misconception: students treat a two-digit number as its digits concatenated, writing the value of "tens digit t, units digit u" as tu (a product) or simply as t + u, instead of 10t + u. The antidote: expand a known number out loud first — 57 is 10·5 + 7 — before introducing letters, and have them rebuild the final number with place value when they check. The same care prevents the parallel error in three-unknown work: forgetting to test the candidate triple against all three equations, not just the two used to find it.