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Lesson 19 – Two-Step Equations

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LESSON 19 / 100  |  19% COMPLETE  |  STAGE II — EQUATIONS
Stage II — Equations

Two-Step Equations

Two layers of wrapping. Two inverse operations. One variable freed.


Why Two Steps?

Most real equations require more than one inverse operation to isolate the variable. A two-step equation has the general form ax + b = c — the variable has been multiplied by a coefficient and had a constant added or subtracted. Both layers must be removed.

ax + b = c

The critical question is: which layer do you remove first? The answer is always the reverse of BODMAS — you undo operations in the opposite order from how they were applied.

1st
Undo + or −
(addition / subtraction)
2nd
Undo × or ÷
(mult / division)
Imagine the variable was dressed first in a multiplication (an inner layer), then in an addition (an outer layer). To undress it, you remove the outer layer first — the addition — then the inner layer — the multiplication. Always work from the outside in.

The Standard Form: ax + b = c

Worked Example A — Solve 3x + 5 = 17
3x + 5 = 17
Step 1 — remove the constant (subtract 5)
3x + 5 − 5 = 17 − 5 subtract 5 from both sides
3x = 12
Step 2 — remove the coefficient (divide by 3)
3x / 3 = 12 / 3 divide both sides by 3

x = 4
Verify
LHS: 3(4) + 5 = 12 + 5 = 17 = RHS ✓
Worked Example B — Solve 4x − 7 = 21
4x − 7 = 21
Step 1 — add 7 to both sides
4x = 28
Step 2 — divide by 4

x = 7
Verify
LHS: 4(7) − 7 = 28 − 7 = 21 = RHS ✓

Negative Coefficients and Negative Results

Worked Example C — Solve −2x + 9 = 3
−2x + 9 = 3
Step 1 — subtract 9
−2x = −6
Step 2 — divide by −2

x = 3 (−6) ÷ (−2) = +3
Verify
LHS: −2(3) + 9 = −6 + 9 = 3 = RHS ✓
Worked Example D — Solve 5 − 3x = 20
5 − 3x = 20
Step 1 — subtract 5 from both sides
−3x = 15
Step 2 — divide by −3

x = −5 15 ÷ (−3) = −5
Verify
LHS: 5 − 3(−5) = 5 + 15 = 20 = RHS ✓

Two-Step Equations with Fractions and Decimals

Worked Example E — Solve x/4 + 3 = 8
x/4 + 3 = 8
Step 1 — subtract 3
x/4 = 5
Step 2 — multiply by 4

x = 20
Verify
LHS: 20/4 + 3 = 5 + 3 = 8 = RHS ✓
Worked Example F — Solve 0.5x − 2.4 = 3.6
Step 1 — add 2.4 to both sides
0.5x = 6.0
Step 2 — divide by 0.5

x = 12 6 ÷ 0.5 = 12
Verify
LHS: 0.5(12) − 2.4 = 6 − 2.4 = 3.6 = RHS ✓

Two-Step Word Problems

Word Problem A — Ayasha thinks of a number, multiplies it by 6, then subtracts 11. The result is 37. Find her number.
1
Let n = the number
2
Equation: 6n − 11 = 37
3
Add 11: 6n = 48
4
Divide by 6: n = 8
5
Ayasha’s number is 8.   Verify: 6(8) − 11 = 48 − 11 = 37 ✓
Word Problem B — A plumber charges a R350 call-out fee plus R180 per hour. A customer’s bill was R1 070. How many hours did the plumber work?
1
Let h = hours worked
2
Equation: 180h + 350 = 1070
3
Subtract 350: 180h = 720
4
Divide by 180: h = 4
5
The plumber worked 4 hours.   Verify: 180(4) + 350 = 720 + 350 = 1 070 ✓

Common Mistakes

Wrong Order — The Classic Error

Dividing before removing the constant is the most frequent two-step mistake:

✗ Wrong — solving 3x + 5 = 17:

Divide by 3 first: x + 5/3 = 17/3 → x = 17/3 − 5/3 = 12/3 = 4

This actually gives the right answer here — but only by accident, and it forces awkward fraction arithmetic. In harder equations it derails entirely. Always remove the constant first.

✓ Correct order: Subtract 5 → divide by 3.

Forgetting to Apply the Operation to Both Sides

 4x − 7 = 21  →  4x = 21   (forgot to add 7 to the RHS)

 4x − 7 = 21  →  4x = 21 + 7 = 28


Practice Set

1. Solve: 2x + 7 = 19
Step 1 — subtract 7: 2x = 12.   Step 2 — divide by 2: x = 6.   Verify: 2(6) + 7 = 19 ✓
2. Solve: 5x − 3 = 27
Add 3: 5x = 30.   Divide by 5: x = 6.   Verify: 5(6) − 3 = 27 ✓
3. Solve: −3x + 4 = 19
Subtract 4: −3x = 15.   Divide by −3: x = −5.   Verify: −3(−5) + 4 = 15 + 4 = 19 ✓
4. Solve: 7 − 4x = −13
Subtract 7: −4x = −20.   Divide by −4: x = 5.   Verify: 7 − 4(5) = 7 − 20 = −13 ✓
5. Solve: x/3 − 6 = 2
Add 6: x/3 = 8.   Multiply by 3: x = 24.   Verify: 24/3 − 6 = 8 − 6 = 2 ✓
6. Solve: 1.2x + 3.6 = 9.6
Subtract 3.6: 1.2x = 6.0.   Divide by 1.2: x = 5.   Verify: 1.2(5) + 3.6 = 6 + 3.6 = 9.6 ✓
7. A school buys identical chairs for each of its 8 classrooms and also one teacher’s desk for R1 200. The total spend is R10 200. Write and solve an equation to find the cost of each chair if every classroom gets 5 chairs.
Total chairs = 8 × 5 = 40.   Let c = cost per chair.   Equation: 40c + 1200 = 10200.   Subtract 1200: 40c = 9000.   Divide by 40: c = R225 per chair.   Verify: 40(225) + 1200 = 9000 + 1200 = 10 200 ✓
8. Challenge: Solve: (2/3)x − 5/6 = 3/2
Add 5/6 to both sides: (2/3)x = 3/2 + 5/6.   LCD = 6: 3/2 = 9/6, so 9/6 + 5/6 = 14/6 = 7/3.   Now (2/3)x = 7/3.   Multiply by 3/2: x = (7/3) × (3/2) = 21/6 = 7/2.   Verify: (2/3)(7/2) − 5/6 = 14/6 − 5/6 = 9/6 = 3/2 ✓

Lesson Checklist

You Can Now
  • Identify the two operations applied to the variable in ax + b = c
  • Apply inverse operations in the correct order: constant first, coefficient second
  • Solve two-step equations with positive and negative coefficients
  • Handle equations where the constant is on the left: b − ax = c
  • Solve two-step equations with fractional and decimal coefficients
  • Translate and solve two-step word problems
  • Verify every solution and identify the two most common errors
Up Next → Lesson 20
Multi-Step Equations

Extend the two-step method to equations requiring simplification first — collecting like terms, distributing, and combining across multiple operations before the variable can be isolated.

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Lesson 18 - One-Step Equations (× and ÷)
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Lesson 20 - Multi-Step Equations