Expanding brackets

Slideshow 11, MathematicsMr Richard Sasaki, Room 307

Objectives• Review algebraic rules learned so

far• Learn how to expand brackets

(known as the Distributive Law)• Expand brackets with unknowns

on the outside

PuzzleTry to figure out the puzzle below!

How much do each of the cars cost?Making algebraic equations may help you solve this problem.

Puzzle - Process

We should consider the cost of each car algebraically.

, , , 𝑎=𝑏𝑐=𝑑×2 +5000

Algebraic equations make the problem look simpler (I hope)!

Puzzle - Result

𝑎=𝑏𝑐=𝑑×2 +5000

So here, we can see doubling is the same thing as adding £5000 to the price.This means that the original price (the cost of and is .£500

0Doubling £5000 gives us the price of and .

£5000

£5000

£10000

£10000

Rules in AlgebraHere are some problems showing rules we know so far.9×𝑥¿9 𝑥𝑥×𝑥¿𝑥2𝑥× 𝑦¿𝑥𝑦𝑥+𝑥¿2 𝑥𝑥+ 𝑦¿𝑥+ 𝑦−6×𝑥¿−6 𝑥𝑥−𝑥¿0𝑥÷ 𝑦¿𝑥𝑦

𝑥÷ 𝑥¿1𝑥2×𝑥3¿𝑥5𝑥5÷ 𝑥3¿𝑥2(𝑥5 )2¿𝑥10(3 𝑥 )2¿9 𝑥2𝑥− 1¿1𝑥𝑥01¿

Expanding BracketsMr. Sasaki lends three students a total of 2 one yen coins and 3 dice each.How could we express this algebraically?

2 𝑥+3 𝑦3 ()Algebraically, how can we show how much of each type of note I handed out?

6 𝑥+9 𝑦These two statements are equal. . Why is this true?

Expanding BracketsLet’s look at the calculation in detail.3 (2 𝑥+3 𝑦 )=¿(3×2𝑥 )+(3×3 𝑦 )

¿6 𝑥+9 𝑦We multiply the number on the outside of the brackets by the numbers on the inside. and remain the same.Let’s try some

examples.2 (2𝑥+ 𝑦 )=¿4 𝑥+2 𝑦3 (𝑥+3 𝑦2 )=¿3 𝑥+9 𝑦25 (2 𝑥−2 𝑦 )=¿10 𝑥−10 𝑦8 𝑥−12 𝑦

4 =¿2 𝑥−3 𝑦

Answers4 𝑥+86 𝑥+3 𝑦8 𝑥−4 𝑦8 𝑥−284 𝑥−6 𝑦−2 𝑥−2 𝑦3 𝑦−3𝑥8 𝑥−4 𝑦3 𝑥+6 𝑦−3𝑧6 𝑥+8 𝑦−2𝑧

Other ExpansionsHow would you calculate this?𝑥 (2 𝑥+3 𝑦 )=¿2 𝑥2+3 𝑥 𝑦

Algebraic values on the outside multiply the inside of the bracket in the same way.Example

2 𝑥 (3 𝑥−2 )=¿2 𝑥×3 𝑥−2𝑥×2Expand

¿6 𝑥2−4 𝑥Note: Expanding brackets is known to some people as the Distributive Law.

Note: Remember…Double means and triple means .

×2 ×3

Answers - Easy6 𝑥 6 𝑥+22 𝑥−2 𝑦 4 𝑥+2 𝑦2 𝑥2−84 𝑥−2 𝑦 2

21 6 𝑦3 𝑥+3 𝑦 6 𝑥−1212𝑥2−36 𝑥2−18𝑥𝑦2 𝑥2+𝑥𝑦 4 𝑥−𝑥𝑦2 𝑥2+2 𝑥𝑦 4 𝑥𝑦−4 𝑦214 𝑥+4 𝑥𝑦 2 𝑥2 𝑦+2𝑥 𝑦2

2 𝑥2+4 𝑥2 𝑥+4 𝑦2

6 𝑥−3 𝑦8 𝑥−8 𝑦14 𝑥+7 𝑦6 𝑥+6 𝑦

Answers - Hard6 𝑥+24 4 𝑥−28𝑥−4 2 𝑥2−4 𝑥4 𝑥𝑦−8 𝑦2 𝑥−26 𝑥𝑦−3 𝑦215𝑥−30 𝑦18 𝑦−6 𝑥3 𝑥2𝑦−3𝑥2 𝑦2

12𝑥2−12𝑥18 𝑥3+63𝑥2

8 𝑥+12 𝑦𝑥2−5

𝑥7+𝑥5 𝑥3 𝑦−𝑥2 𝑦2

11𝑥+22 𝑥3+𝑥2

𝑥4+𝑥2 4 𝑥4+8 𝑥2

15 𝑥3 𝑦2+10𝑥2 𝑦 45𝑎+10𝑏8 𝑥4−202 𝑥+29 𝑥2−153 𝑎 2 𝑥7 𝑦+𝑥2

3 𝑥3+6 𝑥 2 𝑥2−𝑥−1