One unknown, first power — and one routine that always works.
Point 4 of 6 in this lesson: 10.2.4 Solving equations with parentheses
Picture a balance scale that already hangs level: 3x − 5 = 1. You don't know what x weighs yet — that's the whole point — but you know the two pans match. In Lesson 10.1 you learned the two golden rules: do the same thing to both sides and the scale stays level. This lesson turns those rules into a routine — a fixed sequence of moves that takes any linear equation, no matter how tangled, and peels it down to the one line you want: x = something.
We keep the same color habit the whole way down. The unknown is violet, an adding-or-subtracting move is teal, a multiplying-or-dividing move is amber, and the final solution is red. Watch those four colors and you'll always know what kind of step you're looking at.
Before we solve, we have to know what we're allowed to solve. A linear equation in one unknown is the simplest interesting kind of equation. Two words do all the work. One unknown means a single letter — usually x — and only that letter. Linear means the unknown appears only to the first power: just plain x, never x², never 1x, never √x.
Here is the cleanest way to think about it. No matter how messy an equation looks, if it's linear you can always tidy it — gather, combine, move everything to one side — until it reads in the standard form
ax + b = 0, with a ≠ 0.
Here a is the number multiplying the unknown (its coefficient) and b is a plain number. We insist a ≠ 0 because if the x disappeared entirely, there would be nothing left to solve for.
Run a quick test on each one. Is there exactly one letter? Does it appear only as a plain first power? 3x − 5 = 1 passes. 2(x + 1) = x looks busier, but open the bag and it's still just first-power x's — it passes too. But x² = 4 has a square, 2x = 1 hides the unknown under a fraction bar, and √x = 3 has a root. Those three are not linear, and the routine in this lesson is not built for them (you'll meet squares in Stage 11).
An equation is linear in one unknown when a single letter appears, and only ever to the first power. Tidied up, every such equation can be written ax + b = 0 with a ≠ 0.
Here is the single move that makes solving fast. Look at 3x + 5 = 11. We want the x-stuff alone, so the + 5 has to go. The honest, Lesson-10.1 way is to subtract 5 from both sides:
| 3x + 5 = 11 | the equation |
| 3x + 5 − 5 = 11 − 5 | −5 both sides |
| 3x = 11 − 5 | the −5 cancels on the left |
Notice the result: the + 5 vanished from the left and reappeared on the right as − 5. That is the whole trick. So we skip the middle line and just say it out loud:
A term may jump across the equals sign — but it flips its sign as it crosses.
This shortcut has a name: transposition, or "moving terms across." A + 5 becomes − 5; a − 8 becomes + 8; a + 2x on the right becomes − 2x on the left. It is never a new rule — it is always "add or subtract the same thing from both sides," just written in one step instead of two.
The sign flips only when a term crosses the equals sign. Terms that stay on their own side keep their signs. And a term you multiply or divide by (like the 3 in front of x) does not just slide over — moving a coefficient is a different move, coming in the next section.
Pick a term and tap Send it across →. Watch its sign flip, and read the honest "both sides" move it stands for.
Now we have everything we need for a full solve. When the unknown shows up on both sides, the plan is always the same three moves:
1. Send all the x-terms to one side and all the plain numbers to the other (each one flips its sign as it crosses). 2. Combine like terms on each side so you're left with a single x-term and a single number. 3. Divide both sides by the coefficient of x, so its coefficient becomes 1 and x stands alone.
Watch it on 5x − 3 = 2x + 9:
| 5x − 3 = 2x + 9 | the equation |
| 5x − 2x = 9 + 3 | move 2x left, −3 right; each flips sign |
| 3x = 12 | combine like terms |
| x = 4 | ÷ 3 both sides |
And we never trust an answer until we check it. Put x = 4 back into the original equation and make sure both pans really do match:
left: 5(4) − 3 = 20 − 3 = 17 · right: 2(4) + 9 = 8 + 9 = 17 ✓
Both sides equal 17, so x = 4 is right. That last "÷ 3" step is the amber move — dividing both sides by the coefficient — and it's almost always the very last thing you do.
Reveal one line at a time for 5x − 3 = 2x + 9. Say the next step out loud before you tap Next step.
A pair of parentheses is a packed bag. Before you can move the things inside, you have to open the bag — and the tool for that is the distributive law from Stage 4: 2(x − 3) = 2·x − 2·3 = 2x − 6. Multiply the outside number into every term inside. Once the bag is open, you're back to the routine you already know.
Solve 2(x − 3) = x + 4:
| 2(x − 3) = x + 4 | the equation |
| 2x − 6 = x + 4 | distribute the 2 |
| 2x − x = 4 + 6 | move x left, −6 right; each flips |
| x = 10 | combine like terms |
Here the coefficient already came out to 1, so there was no dividing to do. Check in the original: left: 2(10 − 3) = 2·7 = 14; right: 10 + 4 = 14. ✓
The outside number must hit every term, not just the first. 3(x + 2) is 3x + 6, not 3x + 2. And a minus in front of a bag flips every sign inside: −(x − 4) = −x + 4. Think of the minus as a sneaky ×(−1) that touches everything.
Fractions look scary but they're a chore, not a mystery — and there's a one-shot trick to make them disappear. Multiply both sides by the least common multiple (LCM) of the denominators. That single move clears all the fractions at once, because the LCM is a multiple of every denominator (you cleared denominators just like this back in Stage 9).
Solve x+12 − x−13 = 1. The denominators are 2 and 3, so LCM(2, 3) = 6. Multiply every term by 6:
| x+12 − x−13 = 1 | the equation |
| 3(x+1) − 2(x−1) = 6 | × 6 every term |
| 3x + 3 − 2x + 2 = 6 | distribute (mind −2·−1 = +2) |
| x + 5 = 6 | combine like terms |
| x = 1 | move +5 across, it flips to −5 |
Why did 6 turn x+12 into 3(x+1)? Because 6 ÷ 2 = 3. Likewise 6 ÷ 3 = 2, and the lonely 1 on the right becomes 6 × 1 = 6. Now check in the original equation: 1+12 − 1−13 = 22 − 03 = 1 − 0 = 1. ✓
Multiply every term by the LCM — including any plain number sitting by itself, like the 1 here. Forgetting to multiply the lone term is the most common way a fraction problem goes wrong. And when a numerator like x+1 has more than one piece, wrap it in parentheses first: 6 · x+12 = 3(x+1), so the multiplier reaches both pieces.
You've seen me check every answer by substituting it back into the original equation. That's not a formality — it's the safety net that catches a flipped sign or a botched distribution. Always go back to the original, never to a tidied line (a tidied line could carry an early mistake forward).
Most linear equations have exactly one solution. But two surprising things can happen, and both show up when the x-terms cancel each other out. Then the equation collapses to a statement about plain numbers — and that statement is either always false or always true.
Solve 2x + 1 = 2x + 3. Move the 2x across and it cancels:
| 2x + 1 = 2x + 3 | the equation |
| 2x − 2x = 3 − 1 | gather x's left, numbers right |
| 0 = 2 | false — no value of x works |
The x vanished and left 0 = 2, which is simply false. No number can make it true, so this equation has no solution. (And it makes sense: 2x + 1 is always exactly 2 less than 2x + 3 — they can never be equal.)
Solve 2x + 2 = 2(x + 1). Open the bag on the right first:
| 2x + 2 = 2(x + 1) | the equation |
| 2x + 2 = 2x + 2 | distribute the 2 |
| 0 = 0 | true — every x works |
The two sides are identical. Subtract one from the other and you get 0 = 0, which is always true. So every number is a solution — we call such an equation an identity.
Here's the quick way to tell the three cases apart for ax + b = cx + d: compare the x-coefficients first, then the numbers.
| Coefficients | Numbers | Result |
|---|---|---|
| a ≠ c | (any) | one solution |
| a = c | b ≠ d | no solution |
| a = c | b = d | all numbers |
Build your own ax + b = cx + d and the classifier compares a with c, then b with d.
A linear equation in one unknown holds a single letter, only to the first power, and tidies down to ax + b = 0. To solve it, run one routine: clear fractions (× the LCM, every term), open parentheses (distribute to every term, mind a leading minus), move terms across (each flips its sign), combine like terms, then divide by the coefficient so x stands alone. Check by substituting back into the original. If the x-terms cancel, you land on a number statement: false means no solution, true means every number works.
You can now solve any linear equation handed to you. Next you'll learn to build one — turning a story ("a number, doubled, then plus seven…") into an equation, solving it, and answering the real-world question. That's Lesson 10.3 — Putting Linear Equations to Work.
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 the U.S. Common Core standards 7.EE.B.4a (solve word-problem equations of the form px + q = r and p(x + q) = r), 8.EE.C.7a (recognize that a linear equation in one variable has one solution, infinitely many solutions, or no solution, by transforming it into simpler forms), and 8.EE.C.7b (solve linear equations whose solutions require expanding with the distributive property and collecting like terms). The single most common misconception is the sign-flip error when moving a term across the equals sign — students transpose a term but keep its old sign (writing 3x + 5 = 11 → 3x = 11 + 5). The antidote is to keep tying transposition back to its honest meaning: "moving across" is just "add or subtract the same thing from both sides," and a term subtracted from a side that had a plus must arrive on the other side as a minus. Have students briefly write the explicit "−5 from both sides" line until the flip is automatic, and insist on checking every answer in the original equation.