LESSON 17 / 100 | 17% COMPLETE | STAGE II — EQUATIONS
Stage II — Equations
One-Step Equations (+ and −)
The simplest act of solving — one operation undone, one variable freed.
Section 01
The One-Step Principle
A one-step equation requires exactly one inverse operation to isolate the variable. For equations involving addition or subtraction, the inverse operations are straightforward: undo addition with subtraction, and undo subtraction with addition — applied to both sides simultaneously.
Form A — Addition
x + a = b
Inverse: subtract a from both sides
x = b − a
Form B — Subtraction
x − a = b
Inverse: add a to both sides
x = b + a
Section 02
Solving — Addition Equations
Worked Example A — Solve x + 7 = 15
x + 7
=
15
x + 7 − 7
=
15 − 7
subtract 7 from both sides
x
=
8
Verify
LHS: x + 7 = (8) + 7 = 15 = RHS ✓
Worked Example B — Solve x + 14 = 6
x + 14
=
6
x + 14 − 14
=
6 − 14
subtract 14 from both sides
x
=
−8
solution is negative — perfectly valid
Verify
LHS: (−8) + 14 = 6 = RHS ✓
Section 03
Solving — Subtraction Equations
Worked Example C — Solve x − 9 = 4
x − 9
=
4
x − 9 + 9
=
4 + 9
add 9 to both sides
x
=
13
Verify
LHS: (13) − 9 = 4 = RHS ✓
Worked Example D — Solve x − 3 = −11
x − 3
=
−11
x − 3 + 3
=
−11 + 3
add 3 to both sides
x
=
−8
Verify
LHS: (−8) − 3 = −11 = RHS ✓
Section 04
Equations with Decimals and Fractions
The method is identical — the numbers are simply less tidy. Apply the same inverse operation with the same care.
Worked Example E — Solve x + 3.7 = 9.2
x + 3.7 − 3.7
=
9.2 − 3.7
subtract 3.7
x
=
5.5
Worked Example F — Solve x − 2/3 = 5/6
x − 2/3 + 2/3
=
5/6 + 2/3
add 2/3; find LCD = 6
x
=
5/6 + 4/6 = 9/6 = 3/2
Verify
LHS: (3/2) − 2/3 = 9/6 − 4/6 = 5/6 = RHS ✓
Section 05
One-Step Addition/Subtraction Word Problems
Word Problem — After spending R65 on lunch, Thabo had R148 left. How much did he start with?
1
Let x = starting amount (rands)
2
Equation: x − 65 = 148
3
Add 65 to both sides: x = 148 + 65 = 213
4
Answer: Thabo started with R213.
5
Verify: 213 − 65 = 148 ✓
Common Mistakes
| Error | Wrong | Correct |
|---|---|---|
| Applying operation to one side only | x + 7 = 15 → x = 15 | x = 15 − 7 = 8 |
| Using same operation instead of inverse | x + 7 = 15 → x = 15 + 7 = 22 | x = 15 − 7 = 8 |
| Sign error on negative result | x + 14 = 6 → x = 8 | x = 6 − 14 = −8 |
Exercises
Practice Set
1. Solve: x + 9 = 22
Subtract 9: x = 22 − 9 = 13. Verify: 13 + 9 = 22 ✓
2. Solve: x − 15 = 7
Add 15: x = 7 + 15 = 22. Verify: 22 − 15 = 7 ✓
3. Solve: x + 21 = 8
Subtract 21: x = 8 − 21 = −13. Verify: −13 + 21 = 8 ✓
4. Solve: x − 6 = −14
Add 6: x = −14 + 6 = −8. Verify: −8 − 6 = −14 ✓
5. Solve: x + 4.5 = 11.3
Subtract 4.5: x = 11.3 − 4.5 = 6.8. Verify: 6.8 + 4.5 = 11.3 ✓
6. Solve: x − 1/4 = 3/4
Add 1/4: x = 3/4 + 1/4 = 4/4 = 1. Verify: 1 − 1/4 = 3/4 ✓
7. The temperature rose by 11°C to reach 4°C. What was the starting temperature?
Let t = starting temperature. Equation: t + 11 = 4. t = 4 − 11 = −7°C. Verify: −7 + 11 = 4 ✓
8. Challenge: Solve for y: y + 3x = 8x − 5, treating x as a known constant. Simplify your answer.
Subtract 3x from both sides: y = 8x − 5 − 3x = 5x − 5. (This is a solution in terms of x — perfectly valid.)
Summary
Lesson Checklist
You Can Now
- Solve x + a = b by subtracting a from both sides
- Solve x − a = b by adding a to both sides
- Handle negative solutions with confidence
- Apply the method to equations with decimals and fractions
- Verify every solution by substituting back into the original equation
- Translate a one-step word problem into an equation and solve it
- Solve for one variable in terms of another (literal equations preview)
Extend one-step solving to multiplication and division — and handle the special case of negative divisors that flips the sign of the solution.