In 3(x + 2) = 21, the number outside the brackets multiplies everything inside. To solve it, undo the steps in reverse order. First divide both sides by 3 to get x + 2 = 7, then take 2 from both sides to get x = 5. Check by putting 5 back: 3 × 7 = 21. When x is on both sides, expand the brackets and take the smaller x term from both sides.
Explore this idea with maths tools
Maths you can move
Build 8 + 5
Tap two gold counters to fill the ten-frame.
8 in the frame + 5 waiting. The total stays the same.
Words to know
- solve
- find the value of the letter that makes the equation true
- expand
- multiply out the brackets
Worked examples
Solve for x: 3(x + 2) = 21
- 1.Divide both sides by 3: x + 2 = 7.
- 2.Take 2 from both sides: x = 5.
= 5
Solve for x: 2(x + 3) + 4 = 20
- 1.Take 4 from both sides: 2(x + 3) = 16.
- 2.Divide by 2: x + 3 = 8, so x = 5.
= 5
Solve for x: 3(x + 2) = x + 14
- 1.Expand: 3x + 6 = x + 14. Take x from both sides: 2x + 6 = 14.
- 2.Take 6: 2x = 8, so x = 4.
= 4
For parents
Method: Undo in reverse order
What to say: What was done to x last? Undo that first.
Common mistakes:
- Multiplies only the first term in the brackets, so 3(x + 2) becomes 3x + 2.
- Finds the value of the whole bracket and gives that as x.
- Undoes the number inside the brackets with the same operation instead of the opposite one.
- Ignores the number outside the brackets and just undoes the number inside.
- With x on both sides, adds the x terms together instead of taking one from the other.
- With x on both sides, ignores the x term on the right.