An equation hides a number. Somewhere in \(2x + 3 = 11\) there is one value of \(x\) that makes the statement true, and your job is to coax it out of hiding. The remarkable thing is that you do not need to guess. There is a single, dependable idea — keep the equation balanced — that turns every linear equation into a short sequence of obvious moves. Learn that idea once and you can solve them all.
An equation is a balance
Think of the equals sign as the pivot of a balance scale. Whatever sits on the left weighs exactly the same as whatever sits on the right — that is what \(=\) promises. So when we write
\[ x + 5 = 12, \]we are saying the pan on the left, holding an unknown weight \(x\) plus a 5-gram block, balances perfectly against 12 grams on the right.
Here is the golden rule, and it is the whole game: you may do the same operation to both sides, and the balance stays true. Add 3 to both pans, they still balance. Halve both pans, they still balance. As long as you treat the two sides identically, equality survives. Solving is just choosing operations that gradually strip everything away from \(x\) until it stands alone.
In words An equation is a true sentence about numbers. Any operation you apply to one side, apply identically to the other, and the sentence stays true. You are not changing the answer — only changing how the question looks.
One-step equations: undoing a single operation
Every operation has an opposite. Addition is undone by subtraction; multiplication is undone by division. To free \(x\), you simply apply the opposite of whatever is being done to it.
In \(x + 5 = 12\), the number 5 is being added to \(x\). Undo it by subtracting 5 from both sides:
\[ x + 5 - 5 = 12 - 5 \quad\Longrightarrow\quad x = 7. \]In \(3x = 21\), the \(x\) is being multiplied by 3. Undo it by dividing both sides by 3:
\[ \frac{3x}{3} = \frac{21}{3} \quad\Longrightarrow\quad x = 7. \]Notice the pattern: find what is attached to \(x\), and apply the inverse operation to both sides. That single instinct carries you through everything that follows.
Two-step equations: undo in reverse order
Most equations bundle two operations together, in the form \(ax + b = c\). In \(2x + 3 = 11\), the \(x\) is first multiplied by 2, then 3 is added. To peel those layers off, undo them in the reverse order from how they were applied — like taking off your shoes before your socks. Subtract first, then divide.
Watch it happen on a balance. Drag and step through the moves below to see each operation applied to both pans of \(2x + 3 = 11\).
And here is the same reasoning written out as a worked example.
- The \(+3\) is the outer layer, so remove it first. Subtract 3 from both sides: \(2x + 3 - 3 = 11 - 3\), giving \(2x = 8\).
- Now \(x\) is only multiplied by 2. Divide both sides by 2: \(\dfrac{2x}{2} = \dfrac{8}{2}\), giving \(x = 4\).
The hidden number is \(x = 4\). Two clean moves — subtract, then divide — and it was forced into the open.
Tip — undo the outside first. Read the left side as if you were building \(x\) up: "multiply by 2, then add 3." To take it apart, run that backwards: undo the adding before the multiplying. The operation farthest from \(x\) comes off first.
Always check by substituting back
Checking an answer is not optional politeness — it is a free guarantee. Once you think \(x = 4\), put that number back into the original equation and see whether both sides really match.
\[ 2(4) + 3 = 8 + 3 = 11. \checkmark \]The left side equals 11, which is exactly the right side, so \(x = 4\) is correct. If the two sides had disagreed, you would know instantly that a step went wrong — and you could find it without anyone telling you. A thirty-second check catches almost every slip you will ever make.
Variables on both sides
Sometimes the unknown appears on both pans, as in \(5x - 3 = 2x + 9\). The strategy adds just one idea to what you already know: first gather all the \(x\)-terms onto one side, then solve the ordinary two-step equation that remains. Since \(2x\) is the smaller \(x\)-term, subtract it from both sides to keep things positive and tidy.
- Collect the \(x\)-terms. Subtract \(2x\) from both sides: \(5x - 2x - 3 = 9\), which simplifies to \(3x - 3 = 9\).
- Now it is a familiar two-step equation. Add 3 to both sides: \(3x = 12\).
- Divide both sides by 3: \(x = 4\).
So \(x = 4\). Check in the original: the left side is \(5(4) - 3 = 17\) and the right side is \(2(4) + 9 = 17\). Both pans read 17, so the answer holds. \(x = 4\)
The picture below shows what "collecting" really means: \(x\)-terms slide to one side, plain numbers to the other, until the unknown stands alone.
Two special cases
Almost every linear equation has exactly one solution. But two unusual things can happen when you collect terms, and it is worth recognising them so they never throw you.
Sometimes the \(x\)-terms cancel and you are left with a statement that is always true, like \(7 = 7\). That is an identity: every number works, because the two sides were secretly the same expression all along — for instance \(2x + 4 = 2(x + 2)\). Other times the \(x\)-terms cancel and you get something impossible, like \(3 = 5\). That is a contradiction: there is no solution, because no value of \(x\) can make a false number sentence true. If your variable vanishes, read what remains: a true statement means "all numbers," a false one means "none."
Practice
Try each one yourself, then reveal the full solution.
1. Solve \(3x + 5 = 20\).
This is a two-step equation — undo the adding, then the multiplying.
Subtract 5 from both sides: \(3x = 15\). Divide both sides by 3: \(x = 5\).
Check: \(3(5) + 5 = 15 + 5 = 20\). ✓ The solution is \(x = 5\).
2. Solve \(5x - 3 = 2x + 9\).
The variable is on both sides, so collect the \(x\)-terms first.
Subtract \(2x\) from both sides: \(3x - 3 = 9\). Add 3 to both sides: \(3x = 12\). Divide by 3: \(x = 4\).
Check: left side \(5(4) - 3 = 17\), right side \(2(4) + 9 = 17\). ✓ The solution is \(x = 4\).
3. Solve \(4(x - 2) = 12\).
The 4 multiplies the whole bracket, so the quickest move is to divide both sides by 4 right away.
Divide by 4: \(x - 2 = 3\). Add 2 to both sides: \(x = 5\). (Expanding first to \(4x - 8 = 12\), then \(4x = 20\), gives the same answer.)
Check: \(4(5 - 2) = 4 \times 3 = 12\). ✓ The solution is \(x = 5\).