KS3 · Maths

Solving one-step linear equations

Think of a number. Take away 5 and you get 10. You said 15 without a pen? Then you have already solved an equation. Let's make that work every time.

Maths · Algebra

Which move frees x?

Pick an equation, then pick the move that leaves x on its own.

Still to sort

Add a positive number (0)

Something has been taken away from x.

Where the line is: Not 'add a negative number': that is for when something was added to x. Here something was taken away.

Add a negative number (0)

Something has been added to x.

Where the line is: Not 'add a positive number': that is for when something was taken away from x.

Multiply by a whole number (0)

x has been divided by something.

Where the line is: Not 'multiply by a positive fraction': x was divided, so you undo it by multiplying by that whole number.

Multiply by a positive fraction (0)

x has been multiplied by a positive number.

Where the line is: Not 'multiply by a whole number': x was multiplied, so you multiply by a fraction that cancels it, like 1/4 or 7/3.

Multiply by a negative number (0)

x has been multiplied by a negative number.

Where the line is: Not 'multiply by a positive fraction': the number sitting on x is negative, so the move has to be negative too.

7 of 7 still to sort.

In each equation, x is stuck behind one thing. Work out what has been done to x, then pick the move — done to both sides — that undoes it.

Picture it first

A bar model is a picture of an equation: the whole bar on top, the parts it is made of underneath.

Two bar models. Bar one is for a + 6 = 20: a bar of length 20, split into the parts a and 6. Bar two is for 2a = 14: a bar of length 14, split into two equal parts, each one a. Which pair of arrangements gives a on its own for each bar?

Maths · Algebra

Same move, both sides, one line at a time

Step forward to see each new equation and the move that made it. Switch tab for a new equation.

GoalSolve x + 17 = 53

x + 17 = 53

17 has been added to x. To undo that, add the additive inverse of 17: the number that makes 0 when added to 17. That is −17.

Step 1 of 3

17 has been added to x. To undo that, add the additive inverse of 17: the number that makes 0 when added to 17. That is −17.

Maths · Algebra

Where does the working go wrong?

Solve x − 2 = 7.

A student's working — which line goes wrong?

x − 2 = 7
x − 2 + 2 = 7
x = 7

Maths · Algebra

Your turn: fill the gaps

Solve 7x = 2, then solve 0.3x = 6. Check each answer.

Part 1: 7x = 2
missing step

Which line is step 2?

7x × 1/7 = 2 × 1/77x × 7 = 2 × 77x + (−7) = 2 + (−7)

Maths · Algebra

Which of these sounds like you?

Here are four ideas students have about solving equations.

Which idea is closest to what you think right now?

w − 6 = 8, so w = 8 − 6 = 2. The − 6 tells me what to do with the 8.n/4 = 8, so n = 8 ÷ 4 = 2. A division sign means divide.−9m = −45. Dividing by 9 gives −m = −5, and that is my answer.I ask what has been done to the unknown, undo exactly that on both sides, then put my answer back in to check.

How sure are you?

I'm sureI think soI'm guessing

w − 6 = 8, so w = 8 − 6 = 2. The − 6 tells me what to do with the 8.

This is a common misconception

It is easy to see why: the − 6 catches the eye and you want to do something with it. But 8 − 6 = 2 answers a different question. Put w = 2 back in and you get 2 − 6 = −4, not 8. The 6 was taken away from w, so undo that: add 6 to both sides, giving w = 14. Check: 14 − 6 = 8.

n/4 = 8, so n = 8 ÷ 4 = 2. A division sign means divide.

This is a common misconception

n/4 means n ÷ 4, and the division was done to n. To undo a division you multiply: n = 8 × 4 = 32. The tempting 2 fails the check, because 2 ÷ 4 is not 8. Check the real answer: 32 ÷ 4 = 8.

−9m = −45. Dividing by 9 gives −m = −5, and that is my answer.

This is a common misconception

Dividing by 9 was a good move, but −m = −5 still has a minus sign in front of the unknown, so m is not on its own yet. Multiply both sides by −1 and you get m = 5. Check: −9 × 5 = −45.

I ask what has been done to the unknown, undo exactly that on both sides, then put my answer back in to check.

This matches the evidence

That habit solves every one-step equation: additive or multiplicative, with negatives, fractions or decimals. The check is the part that makes you independent. It tells you whether you are right without anyone marking it.

Do not read an equation as a sum to work out. Ask what has been done to the unknown, undo it on both sides, then check by substituting.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Undo what was done to x, on both sides, then check.

What you need to know

A solution is the value of the unknown that keeps both sides equal. Finding it is called solving, and it means getting the unknown on its own.
Ask: what has been done to x? Then do the opposite to both sides of the equation. An addition or subtraction is undone with an addition; a multiplication or division with a multiplication.
A solution does not have to be a whole number, and it does not matter which side the unknown is on.
Always check: put your answer into the original equation and see whether both sides match.

The big picture

To solve a one-step equation, ask what has been done to the unknown and undo exactly that on both sides, so the unknown is left on its own. Then check by substituting your answer back in.

Key points

1Solving means isolating the unknown: x ends up on its own.

2Whatever you do to one side of an equation, do to the other, so equality is kept.

3x − 5 = 10 has the solution x = 15, not 5. x/3 = 12 has the solution x = 36, not 4.

4If −x is left, multiply both sides by −1: only x on its own is a solution.

5Substituting the answer back in tells you at once whether it is right.

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Worked example

Problem

Solve 2/5 x = 8, then check.

⚠ Watch out

Reading the equation as a sum. x − 5 = 10 is not 10 − 5 = 5, since 5 − 5 is 0. x/3 = 12 is not 12 ÷ 3 = 4, since 4 ÷ 3 is not 12. Undo what was done to x, on both sides.

🧠

Memory hook

Undo it, on both sides, then check. 5 − 5 is not 10, and 4 ÷ 3 is not 12.

✓

Check yourself

Without looking back: for s + 31 = 50 and for s/6 = 5, say which move you do to both sides each time. Then solve both and check your answers.

Flashcards

(13)

What is a solution to an equation?

The value of the unknown that keeps both sides of the equation equal.

What does it mean to solve an equation, and to isolate the unknown?

Solving means finding the value of the unknown. To do it, isolate the unknown: get it on its own, like x = 36, so its value can be seen.

Does it matter which side the unknown is on?

No. 8 = b says the same as b = 8, so the solution is b = 8.

Additive inverse

The number you add to a number to make 0. The additive inverse of 7 is −7, and the additive inverse of −17 is 17.

Zero pair

A number and its additive inverse, like x and −x, or 0.3 and −0.3. Added together they make 0.

How do you solve x − 21 = −6?

The constant is −21, so add its additive inverse, +21, to both sides: x = 15. Check: 15 − 21 = −6.

Reciprocal

The multiplicative inverse of a non-zero number: a number times its reciprocal is 1. The reciprocal of 3/7 is 7/3.

How do you solve 5s = 30?

Multiply both sides by the reciprocal of 5, which is 1/5 (the same as dividing both sides by 5): s = 6.

How do you solve h/4 = 20?

h/4 means h ÷ 4, so multiply both sides by 4: h = 80. Dividing 20 by 4 gives 5, but 5 ÷ 4 is not 20.

Are 2/7 and 7/2 the same? Which solves 7x = 2?

They are different values. 7x = 2 gives x = 2/7 (a seventh of 2), while 2x = 7 gives x = 7/2.

What do you do with a negative coefficient, as in −5x = 35?

Isolate x itself, not −x: divide both sides by −5 (or multiply by −1, then divide by 5). x = −7.

How do you check a solution?

Substitute it into the original equation. If the two sides do not match, the solution is wrong.

Can one additive step solve 2x + 7 = 16?

No. Adding 3, −x or −7 to both sides keeps equality but does not show what x is. It needs more than one step.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Which move frees x?

Same move, both sides, one line at a time

Where does the working go wrong?

Your turn: fill the gaps

Which of these sounds like you?

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