An equation with a bracket is expanded first, then solved with the usual moves
Expand the bracket first
In 3(x + 2) = 15, the letter is inside a bracket, and the bracket is multiplied by 3. The first move is to expand the bracket, so that the equation has a shape you already know how to solve.
The 3 multiplies everything inside the bracket: 3 × x = 3x and 3 × 2 = 6. So 3(x + 2) = 3x + 6, and the equation becomes 3x + 6 = 15.
3(x + 2) is a rectangle 3 tall and x + 2 long. It is a 3 by x piece and a 3 by 2 piece, so it is 3x + 6.
Then solve as usual
Now undo the operations around x, doing the same to both sides each time. Subtract 6 from both sides: 3x + 6 − 6 = 15 − 6, so 3x = 9. Then divide both sides by 3: x = 3.
Check the answer in the equation you started with, the one with the bracket: 3(3 + 2) = 3 × 5 = 15. Both sides agree, so x = 3 is right.
The 2 must be multiplied too
The usual mistake is to multiply only the x and let the 2 through untouched: 3(x + 2) = 15 becomes 3x + 2 = 15. That leads to 3x = 13 and , and the check fails: , not 15.
In the rectangle, the 3 runs along the whole of x + 2, so it covers the 2 as well as the x. The piece 3 by 2 is 6, and it is part of the left side of the equation.
A minus sign in front
The number in front can be negative. Then it multiplies each term inside, and the signs follow the rules for multiplying negative numbers. In −2(x − 4), −2 × x = −2x and −2 × (−4) = +8, so −2(x − 4) = −2x + 8.
So 5x − 2(x − 4) = 20 becomes 5x − 2x + 8 = 20. Collect the x terms: 3x + 8 = 20. Subtract 8 from both sides: 3x = 12, and x = 4. Check: 5 × 4 − 2(4 − 4) = 20 − 0 = 20.
Writing −2(x − 4) as −2x − 8 is the slip to watch for: a negative times a negative is positive, so the last term is +8.
Two brackets
When an equation has more than one bracket, expand each one, then collect like terms before solving. In 3(x + 1) + 2(x + 4) = 21, the brackets give 3x + 3 and 2x + 8. Collect them: 5x + 11 = 21. Subtract 11 from both sides: 5x = 10. Divide both sides by 5: x = 2. Check: 3 × 3 + 2 × 6 = 9 + 12 = 21.
Negative numbers and linear equations · A Rectangle with Sides Written in Terms of x
QuestionA rectangular picture frame is (2x + 5) cm long and (x − 1) cm wide. A strip of wood exactly 50 cm long goes once round its edge. (a) Find the value of x. (b) Find the area of the rectangle.
1.The perimeter is twice the length plus twice the width, and it is 50 cm: 2(2x + 5) + 2(x − 1) = 50.
Round the edge go the length, the width, the length and the width: 2(2x + 5) + 2(x − 1) = 50. 2.Expand the brackets: 4x + 10 + 2x − 2 = 50. Collect like terms: 6x + 8 = 50.
Expand the brackets: 4x + 10 + 2x − 2 = 50. Collect like terms: six x's and 10 − 2 = 8, so 6x + 8 = 50. 3.Subtract 8 from both sides: 6x = 42. Divide both sides by 6: x = 7.
Subtract 8 from both sides: 6x = 42. Divide both sides by 6: x = 7. 4.(a) x = 7. The length is 2 × 7 + 5 = 19 cm and the width is 7 − 1 = 6 cm. Both sides are positive lengths, so this value of x fits the rectangle.
(a) x = 7, so the rectangle is 2 × 7 + 5 = 19 cm long and 7 − 1 = 6 cm wide, both positive lengths. 5.The area is the length times the width: 19 × 6 = 114.
The area is the length times the width: 19 squares in each row, 6 rows, 19 × 6. 6.(b) The area is 114 cm2. Check: the perimeter is 2 × (19 + 6) = 2 × 25 = 50 cm.
(b) The area is 19 × 6 = 114 cm2. Check: the perimeter is 2 × (19 + 6) = 50 cm.
Answer: (a) x = 7; (b) 114 cm2
Watch out for
- Adding one length and one width only, (2x + 5) + (x − 1) = 50. A rectangle has two lengths and two widths, so the sum of one of each is half the perimeter, 25 cm.
- Expanding 2(x − 1) as 2x − 1. The 2 multiplies both terms inside the bracket, so 2(x − 1) = 2x − 2.
Your turn
1.Solve 2(x + 6) = 20