Two unknowns are one too many — so make one disappear.
Point 3 of 5 in this lesson: 10.5.3 Elimination by adding or subtracting
In Lesson 10.4 you met a system: two equations, two unknowns, sharing one secret pair (x, y) that fits both. Drawing two lines and reading off the crossing point is honest, but a graph rarely lands you on a tidy answer like (2, 3) — it lands you near it. This lesson hands you the algebra that gets the exact pair every time. The whole trick has one name: elimination — get rid of one unknown so you are back to the one-unknown equations you already conquered in Lesson 10.2.
Keep the color habit from the whole stage: the unknowns x and y are violet, equation one is teal, equation two is amber, and the solution pair is red. When two terms cancel, watch them turn to dust.
One unknown you can handle. Two unknowns crowd each other: in a single equation like 2x + y = 7, every choice of x just hands you a matching y, so there is no single answer. A second equation pins things down — but only if you can untangle the two letters.
Here is the one idea that runs through this entire lesson. If you can knock out one of the unknowns, you are left with one equation in one unknown — and that is a problem you already know how to finish. Everything below is just a different way to make a letter disappear.
There are two everyday ways to eliminate a letter. You can substitute — solve one equation for a letter and pour that whole expression into the other, so only one letter is left standing. Or you can add or subtract the equations themselves — line them up and combine them so a matching pair of terms cancels. Same goal, two roads. The next sections walk each one, slowly.
Two unknowns are one too many. Every method in this lesson does the same thing: eliminate one unknown to get a single one-unknown equation, solve it, then back-substitute to find the other.
Suppose one equation already tells you what a letter is. Take the system
y = 2x − 1 and 3x + y = 9.
The first equation says y is exactly the same thing as 2x − 1. So anywhere we see y, we are allowed to write (2x − 1) instead — they are equal, so the swap changes nothing. Pour that expression into the second equation and the y is gone:
| 3x + y = 9 | equation two |
| 3x + (2x − 1) = 9 | replace y with 2x − 1 |
| 5x − 1 = 9 | combine like terms: 3x + 2x = 5x |
| 5x = 10 | add 1 to both sides |
| x = 2 | divide both sides by 5 |
Now climb back up. We know x = 2, and the first equation hands us y directly:
| y = 2x − 1 | equation one |
| y = 2(2) − 1 | put x = 2 in |
| y = 3 | 4 − 1 = 3 |
The secret pair is (2, 3). Always check it in both originals, because a single check can hide an arithmetic slip:
2(2) − 1 = 3 ✓ 3(2) + 3 = 9 ✓
Substitute the whole expression, kept inside parentheses: write 3x + (2x − 1), not 3x + 2x − 1 dropped in by hand. The parentheses keep the sign of every term safe when the expression carries a subtraction or a negative coefficient.
Step through y = 2x − 1 and 3x + y = 9. Each tap uncovers the next move; nothing is skipped.
Sometimes you do not need to isolate anything — the equations are already lined up for a clean cancel. Look at
x + y = 7 and x − y = 1.
The y term is +y in the first equation and −y in the second. Those are opposites. If we add the two equations — left side to left side, right side to right side — the y terms add to zero and disappear:
| (x + y) + (x − y) = 7 + 1 | add the two equations |
| 2x = 8 | the y terms cancel |
| x = 4 | divide both sides by 2 |
Back-substitute x = 4 into either original. Using x + y = 7: 4 + y = 7, so y = 3. The solution is (4, 3). Check: 4 + 3 = 7 ✓ and 4 − 3 = 1 ✓.
When do you add and when do you subtract? It depends on the matching pair:
| The matching terms are… | Move | Why it cancels |
|---|---|---|
| opposite (e.g. +y and −y) | ADD | +y + (−y) = 0 |
| equal (e.g. 3x and 3x) | SUBTRACT | 3x − 3x = 0 |
Subtracting an equation means subtracting every term on both sides — flip the sign of each one. From 3x + y = 9 minus 3x − 2y = 1 you get (y − (−2y)) = 3y on the left and (9 − 1) = 8 on the right, so 3y = 8. Forgetting to flip the sign of the right side is the classic slip.
Pick which letter to kill. The widget decides add-versus-subtract from the signs, then shows the cancel and finishes the solve. Every pair lands on whole numbers.
Adding works only when a pair already matches or is opposite. Most systems are not so polite. Consider
2x + 3y = 12 and 3x − 2y = 5.
No letter matches: the x coefficients are 2 and 3; the y coefficients are 3 and −2. But we are allowed to multiply a whole equation by a number — both sides, every term — and it still describes the same line. So we manufacture a match. Let us eliminate y: we want the y coefficients to become opposites. The numbers are 3 and −2; their least common multiple is 6. Multiply the first equation by 2 and the second by 3:
| 2x + 3y = 12 × 2 | scale equation one |
| 4x + 6y = 24 | every term doubled |
| 3x − 2y = 5 × 3 | scale equation two |
| 9x − 6y = 15 | every term tripled |
| 13x = 39 | add: +6y and −6y cancel |
| x = 3 | divide both sides by 13 |
Climb back up with x = 3, into the first original equation:
| 2(3) + 3y = 12 | put x = 3 into equation one |
| 6 + 3y = 12 | 2 × 3 = 6 |
| 3y = 6 | subtract 6 from both sides |
| y = 2 | divide both sides by 3 |
So (3, 2). Check both originals: 2(3) + 3(2) = 6 + 6 = 12 ✓ and 3(3) − 2(2) = 9 − 4 = 5 ✓. Perfect.
To cancel a letter, make its two coefficients match in size. Find the least common multiple of the coefficients, then multiply each equation by whatever turns its coefficient into that LCM. Here the y coefficients were 3 and 2, LCM 6, so the multipliers were 6 ÷ 3 = 2 and 6 ÷ 2 = 3. (You could instead have targeted x: coefficients 2 and 3, LCM 6, multipliers 3 and 2 — same amount of work, same answer.)
Choose which letter to cancel. Watch the widget pick multipliers, scale both equations, and check whether the new coefficients are opposite (add) or equal (subtract) before it cancels.
You now own three moves: substitute, add-or-subtract, and scale-then-add. They all reach the same answer, so the only question is which is least work for the system in front of you.
| If the system looks like… | Reach for… |
|---|---|
| a letter is already alone, like y = …, or has coefficient 1 | substitution |
| a pair of coefficients already match or are opposite | add / subtract |
| nothing lines up | scale one (or both) equations, then add / subtract |
There is no wrong choice — only faster and slower ones. If y sits alone on one side, substitution is a gift; do not multiply equations for no reason. If you see 5x in one equation and 5x in the other, subtracting kills x in a single line; do not bother isolating anything.
Whatever path you take, the last step never changes: put the pair back into BOTH original equations. A solution must satisfy both — that is what makes it a solution to the system. If it fits one and not the other, you have an arithmetic slip to hunt down, not an answer.
Pick the method that needs the fewest moves, but always verify in both originals. The crossing point of the two lines is the one pair that makes both equations true at once.
Choose a system and a method. The widget solves it step by step and lands a red dot on the crossing point. Both methods reach the same pair — try each.
Two unknowns are one too many, so make one disappear. Substitution pours one equation's value for a letter into the other; adding or subtracting cancels a matching pair directly; and when nothing matches, scale an equation first to manufacture a pair — equal coefficients call for subtracting, opposite ones for adding. Each road shrinks the system to a single one-unknown equation you already know how to solve, then you back-substitute for the partner and check the pair in both originals. The answer is the one point where the two lines cross.
Elimination is a tool; Lesson 10.6 aims it at real problems — mixtures, ages, speeds, and money — and then stacks a third equation on top, eliminating down from three unknowns to two to one. Same idea, one more floor.
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 8.EE.C.8b (solve systems of two linear equations algebraically — by substitution and by elimination — and estimate solutions by graphing) and A.REI.C.5–6 (solving a system exactly, including by replacing one equation with a sum of it and a multiple of the other). The single most common misconception is the sign error when subtracting one equation from another: students subtract the left sides but forget to flip every sign on the right, or they drop the negative on the eliminated term. The antidote is to rewrite subtraction as adding the opposite — multiply the whole second equation by −1 first, then add — and to insist on the non-negotiable last step: substitute the pair back into BOTH original equations. A pair that satisfies only one equation is not a solution to the system.