Equations with Brackets

Order the same combo for a group — a main of unknown price plus a few fixed extras, several times over — and the total bill packs the unknown inside a bracket. That is the shape of equation this page cracks.

Sometimes the unknown is locked inside a bracket, with a number standing guard outside it:

3(x + 4) = 27

You can't get at x in one move — it is trapped inside, and everything in the bracket is multiplied by the 3. The fix is a single extra step: knock the bracket down first. Expand it by multiplying the 3 into both terms inside:

3(x + 4) = 3x + 12

The bracket has vanished. What is left is an ordinary two-step equation — something you already know how to finish:

3x + 12 = 27

Finish it off

From here, undo in reverse: first subtract the 12 that was added, then undo the multiplication by 3.

3x + 12 = 27 \;\Rightarrow\; 3x = 15 \;\Rightarrow\; x = 5

Check it in the original: 3(5 + 4) = 3 \times 9 = 27 — spot on. So the whole recipe is: expand the bracket, then solve the two-step equation. One small new move unlocks a huge family of problems.

The distributive law — the engine behind it

Expanding a bracket is really one famous rule wearing a hat: the distributive law. It says a multiplier outside a bracket gets shared out — distributed — across everything inside:

a(b + c) = ab + ac

You already use this without noticing. To work out 6 \times 99 in your head, you don't grind through long multiplication — you think of 99 as (100 - 1) and distribute:

6 \times 99 = 6(100 - 1) = 600 - 6 = 594

Same rule, whether the bracket hides a spare 1 or a mystery x.

See it solved

Step through the solution: first the bracket is expanded, then the two-step equation is solved one undo at a time.

A story hiding a bracket

Brackets show up naturally whenever the same bundle repeats. Suppose you fill party bags, each with an unknown number of sweets x plus a fixed 4 stickers. You make 3 identical bags and use up 27 items in total. How many sweets went in each bag?

"3 lots of (x + 4), total 27" is written straight down as a bracket equation:

3(x + 4) = 27

Expand to 3x + 12 = 27, then solve: 3x = 15, so x = 5. Five sweets per bag. The bracket captured "the same bundle, three times over" in a single clean line — that is what brackets are for.

Worked example — brackets on both sides

Brackets don't only appear alone. Here x turns up on both sides:

2(x + 1) = x + 8

Expand the left-hand bracket first:

2x + 2 = x + 8

Now gather the x's onto one side. Subtract x from both sides, then finish as a two-step equation:

2x + 2 - x = x + 8 - x \;\Rightarrow\; x + 2 = 8 \;\Rightarrow\; x = 6

Check: 2(6 + 1) = 14 and 6 + 8 = 14. Both sides agree, so x = 6.

Worked example — when expanding leaves a negative term

Expanding sometimes produces a subtraction. Watch the sign carefully:

3(x - 5) = 6 3x - 15 = 6

The bracket gave -15. Now undo it — add 15, then divide by 3:

3x - 15 + 15 = 6 + 15 \;\Rightarrow\; 3x = 21 \;\Rightarrow\; x = 7

Check: 3(7 - 5) = 3 \times 2 = 6. Correct — x = 7.

See it explained

Sal Khan distributes across a bracket, then solves the equation that's left.

The number-one error with brackets, and the sign trap that follows it:

Usually — but not always. For 3(x + 4) = 27 there is a sneaky shortcut: since the whole left side is 3 times the bracket, and the right side is 27, you could divide both sides by 3 first. That gives x + 4 = 9 in one move, so x = 5 — no expanding needed! This works nicely when the number outside divides the right-hand side cleanly.

But the moment the right side doesn't divide evenly, or there are terms on both sides, dividing first turns into fractions and misery. That is why "expand first" is the reliable all-purpose method you should learn cold — the clever shortcut is a bonus for when the numbers are kind.

A bracket lets a single equation stand in for a whole family of situations. Say a phone plan charges a £4 monthly fee plus the same amount for each of x extra add-ons, and three friends all take the identical plan — the total bill is 3(x + 4). Change the story to chemistry and the very same distributive law balances the atoms on each side of a reaction. Expanding brackets isn't a niche classroom chore: it is the move that turns "the same thing, several times over" into one tidy equation you can actually solve — which is why it shows up absolutely everywhere.