4 Solving simple equations with one unknown
Consider the equation 4x = 24. As it is written x is the unknown and to solve this equation you need to find the value of x which makes the statement 4x = 24 true.
You may be able to see this straight away, but if not, divide both sides of the equation by 4, to get an answer for x.
The solution is, therefore, x = 6.
You can check the accuracy of your answer by substituting the value for x back into the original equation: 4 × 6 = 24.
Try this examples for yourself now.
Activity 4 Solving equations
Now solve the following equations for x.
- 3x = 6
- 7x = 21
- 8x = 32
- x + 3 = 9
- x + 6 = 7
- x + 7 = 11
Answer
- 3x = 6 divide both sides by 3: x = 2
- 7x = 21 divide both sides by 7: x = 3
- 8x = 32 divide both sides by 8: x = 4
- x + 3 = 9 subtract 3 from both sides : x = 6
- x + 6 = 7 subtract 6 from both sides : x = 1
- x + 7 = 11 subtract 7 from both sides : x = 4
Now you’ve warmed up try some slightly more complex equations, which involve rearranging equations and multiplying out of brackets.
Activity 5 More complex equations
Now solve the following equations for m.
- 2x – 1 = 7
- 5x – 8 = 2
- 3x + 8 = 5
- 3x – 8 = 5x – 20
- 3(x + 1) = 9
- 23 – x = x + 11
- 2(x – 3) – (x – 2) = 5
Answer
| Equation | Step 1 | Step 2 | Solution | |
|---|---|---|---|---|
| a. | 2x – 1 = 7 | 2x – 8 | x = 4 | |
| b. | 5x – 8 = 2 | 5x = 10 | x = 2 | |
| c. | 3x + 8 = 5 | 3x = –3 | x = –1 | |
| d. | 3x – 8 = 5x – 20 | –8 = 5x – 3x – 20 | 12 = 2x | x = 6 |
| e. | 3(x + 1) = 9 | 3x + 3 = 9 | 3x = 6 | x = 2 |
| f. | 23 – x = x + 11 | 23 = 2x + 11 | 2x = 12 | x = 6 |
| g. | 2(x – 3) – (x – 2) = 5 | 2x – 6 – x + 2 = 5 | x – 4 = 5 | x = 9 |
| h. | 4x = 60 | x = 15 | ||
| i. | x = 1.2 | |||
| j. | 5m + 3m = 30 | m = 3.75 | ||
| k. | 7x – 4x = 56 | x = 18.7 (to 1 decimal place) |
Rather than solving just one equation in the next section you will learn about pairs of equations with two unknowns.