3
Brackets and the distributive property: adding expressions and evaluating
Mathematics Prep 1 Term 1, Unit 2, Lesson 1
GoalRemove brackets with the distributive property, add expressions and find the value for a given number.
New words
distributive property · a(b + c) = ab + acexpand · remove brackets by multiplying every term insidesubstitute · replace a letter by a numbervalue · the number an expression gives after substituting
Explanation
The distributive property says a(b + c) = ab + ac and a(b − c) = ab − ac. Multiply the number outside by every term inside: 3(x − 4) = 3x − 12. A minus sign before a bracket changes every sign inside: −(x − y) = −x + y, and −(−x) = x. To add or subtract expressions, remove the brackets and collect like terms: (3x + 5) + (2x − 7) = 5x − 2. To find a value, substitute the number for the letter: for 3x − 5 with x = 4, we get 3 × 4 − 5 = 7.
ƒFormula
a(b + c) = ab + ac
Also a(b − c) = ab − ac and −(x − y) = −x + y.
Worked examples
Expand 3(x − 4): multiply 3 × x = 3x and 3 × 4 = 12, so 3x − 12. Expand 2(3a + 5) = 6a + 10. Check with a = 1: 2(3 + 5) = 16 and 6 + 10 = 16.
A minus before a bracket. −(x − y) = −x + y. And 5 − (2x − 3) = 5 − 2x + 3 = 8 − 2x. Subtract expressions: (4x + 1) − (x − 3) = 4x + 1 − x + 3 = 3x + 4.
| x | 4x + 4 |
|---|---|
| 5 | 24 |
| 10 | 44 |
| 15 | 64 |
Class activity
Check by numbers. With a grown-up, pick x = 2. Work out 3(x + 4) and also 3x + 12. Do they agree? Try x = 5. Then try a wrong expansion, 3x + 4, and see that it disagrees.
Practice
1
Expand 5(2x − 3).
10x − 15.
2
Expand −(3x − 2).
−3x + 2.
3
Add (4x + 3) and (x − 8).
5x − 5.
4
Find the value of 3x − 5 when x = 6.
3 × 6 − 5 = 13.
5
Seif writes 3(x − 4) = 3x − 4. What is wrong? Check with x = 5.
He forgot to multiply the 4. 3(x − 4) = 3x − 12. With x = 5: 3 × 1 = 3 and 15 − 12 = 3, but 15 − 4 = 11.