Mathematics Standard • Year 11 • Module 1 • Lesson 2

Solving One-Step and Two-Step Equations

Build fluency using inverse operations to solve one-step, two-step and bracketed equations, checking by substitution.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the inverse operation for each.

"Add 7" undoes with ____________. "Multiply by 4" undoes with ____________.

"Subtract 12" undoes with ____________. "Divide by 5" undoes with ____________.

Q1.2 In the equation 3x + 5 = 26, which operation is undone first when solving? Circle one: add 5 / subtract 5 / multiply by 3 / divide by 3.

Q1.3 A solution to an equation is a value that ____________ the equation true. Write the word that fills the blank: ____________.

Stuck? Revisit lesson § Two-Step Equations — undo the outside operation (the +5) before the multiplication.

2. Worked example — solve and check 3x + 5 = 26

Follow each balancing step. Every move keeps both sides equal.

Problem. Solve 3x + 5 = 26 using inverse operations, then check by substitution.

Step 1 — Undo the +5 first (subtract 5 from both sides).

3x + 5 − 5 = 26 − 5 ⇒ 3x = 21

Reason: the +5 was the last operation built onto x, so it is the first to be undone.

Step 2 — Undo the ×3 (divide both sides by 3).

3x / 3 = 21 / 3 ⇒ x = 7

Reason: division is the inverse of multiplication; both sides stay equal.

Step 3 — Check by substitution.

3(7) + 5 = 21 + 5 = 26 ✓

Reason: substituting x = 7 returns the original 26, confirming the solution.

Conclusion. The solution is x = 7.

3. Faded example — fill in the missing steps

Solve 4x − 9 = 23 and check by substitution. Fill in each blank. 4 marks

Step 1 — Undo the −9 first:

4x − 9 + ____ = 23 + ____ ⇒ 4x = ____________

Step 2 — Undo the ×4 (divide both sides by ____):

x = ____________

Step 3 — Check by substitution:

4(____) − 9 = ____________ − 9 = ____________ ✓

Conclusion. The solution is x = ____________.

Stuck? Revisit lesson § Worked Example 2 — Solve and check 3x + 5 = 26.

4. Graduated practice — solving equations

Show every balancing step. Write a check by substitution for any answer marked with a (✓) tag.

Foundation — one-step equations (4 questions)

Q	Equation	Solution
4.1 1	y + 8 = 23	y = ________
4.2 1	y − 6 = 15	y = ________
4.3 1	5a = 45	a = ________
4.4 1	x / 4 = 6	x = ________

Standard — two-step equations (6 questions)

Show subtraction/addition first, then multiplication/division.

4.5 Solve 2m + 7 = 31 and check. (✓) 2 marks

4.6 Solve 5x − 4 = 26. 2 marks

4.7 Solve 6x + 9 = 51. 2 marks

4.8 Solve 4x + 9 = 37 and check by substitution. (✓) 2 marks

4.9 Solve x/3 + 5 = 11. 2 marks

4.10 Solve 7x − 12 = 23. 2 marks

Extension — bracketed and negative coefficient (2 questions)

4.11 Solve 4(p − 3) = 28 in two different ways: (a) divide both sides by 4 first; (b) expand the brackets first. Show both methods give the same answer. 3 marks

4.12 Solve 17 − 2x = 5 and check by substitution. 3 marks

Stuck on 4.12? Subtract 17 from both sides first to get −2x = −12, then divide by −2 (a negative ÷ negative = positive).

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (y + 8 = 23) I subtracted 8 from both sides to get y = 15.

For 4.3 (5a = 45) I divided both sides by 5, not just one side.

For 4.4 (x/4 = 6) I multiplied both sides by 4 (the inverse of dividing by 4) to get x = 24.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Inverse operations

Add 7 ↔ subtract 7. Multiply by 4 ↔ divide by 4. Subtract 12 ↔ add 12. Divide by 5 ↔ multiply by 5.

Q1.2 — Order in 3x + 5 = 26

Subtract 5 first (undo the addition before the multiplication).

Q1.3 — Definition

A solution is a value that makes the equation true.

Q3 — Faded example (4x − 9 = 23)

Step 1: 4x − 9 + 9 = 23 + 9 ⇒ 4x = 32.
Step 2: divide by 4 ⇒ x = 8.
Step 3: Check 4(8) − 9 = 32 − 9 = 23 ✓.
Conclusion: x = 8.

Q4.1 — y + 8 = 23

Subtract 8: y = 15.

Q4.2 — y − 6 = 15

Add 6: y = 21.

Q4.3 — 5a = 45

Divide by 5: a = 9.

Q4.4 — x/4 = 6

Multiply both sides by 4: x = 24.

Q4.5 — 2m + 7 = 31

Subtract 7: 2m = 24. Divide by 2: m = 12. Check: 2(12) + 7 = 24 + 7 = 31 ✓.

Q4.6 — 5x − 4 = 26

Add 4: 5x = 30. Divide by 5: x = 6.

Q4.7 — 6x + 9 = 51

Subtract 9: 6x = 42. Divide by 6: x = 7.

Q4.8 — 4x + 9 = 37

Subtract 9: 4x = 28. Divide by 4: x = 7. Check: 4(7) + 9 = 28 + 9 = 37 ✓.

Q4.9 — x/3 + 5 = 11

Subtract 5: x/3 = 6. Multiply by 3: x = 18.

Q4.10 — 7x − 12 = 23

Add 12: 7x = 35. Divide by 7: x = 5.

Q4.11 — 4(p − 3) = 28 (two methods)

(a) Divide first. 4(p − 3) = 28 ⇒ p − 3 = 7 ⇒ p = 10.
(b) Expand first. 4p − 12 = 28 ⇒ 4p = 40 ⇒ p = 10.
Both give p = 10. (Method a is usually faster when the outside number divides the right side cleanly.)

Q4.12 — 17 − 2x = 5

Subtract 17: −2x = −12. Divide both sides by −2: x = 6. Check: 17 − 2(6) = 17 − 12 = 5 ✓. (Trap: dropping the negative sign and getting x = −6 — substitute and you'll get 17 + 12 = 29, not 5.)