L4 · Finding an unknown in a one-operation equation (either side); problems with equations

Learning goal

“I can find the unknown in an equation with one operation - whether the unknown is on the left or the right - by undoing the operation, and I can check my answer by putting it back into the equation. I can write and solve an equation for a problem.”

Solve — to find the value of the unknown that makes an equation true.

Solution — that value itself: in □ + 7 = 18, the solution is □ = 11.

Undo — to reverse an operation: adding is undone by subtracting, dividing is undone by multiplying, and the other way round.

Check — to put the solution back into the original equation and confirm both sides have the same value.

Worked example

Solve □ - 6 = 11. Step 1: Name what happened to the box: 6 was SUBTRACTED from it. Step 2: Undo it - add 6 back: □ = 11 + 6 = 17. Step 3: Check in the ORIGINAL equation: 17 - 6 = 11. Second form: 20 = □ + 8. The unknown is on the right; undo the +8 by subtracting 8 from both sides: □ = 20 - 8 = 12. Check: 12 + 8 = 20. Balance view, in words: undoing is taking the same weight off both pans - or putting the same weight back on both pans - until the box sits alone.

Check yourself — explain it

In □ - 6 = 11, why do we ADD 6 to find the box?

Paired problem — example still visible

Solve and check: □ + 9 = 21 -> □ = 12; check: 12 + 9 = 21

Faded completion — last step is yours

"16 = □ - 4: undo by adding 4, so □ = 16 + 4 = ___" -> 20

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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