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Expanding brackets and substitution

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Presentation on theme: "Expanding brackets and substitution"β€” Presentation transcript:

1 Expanding brackets and substitution
Slideshow 8, Room 307 Mathematics, Mr Richard Sasaki

2 Objectives Recall how to substitute into expressions
Practice expanding brackets Expand brackets before substituting into expressions

3 Substitution To substitute, we swap algebraic symbols for numbers so we can get a numeric answer. Example Calculate π‘₯+𝑦 when π‘₯=4 and 𝑦=7. Try the short worksheet. π‘₯+𝑦 = 4+7 = 11 Usual algebraic rules for expressions apply.

4 Substitution Answers 9 2 56 144 75 32 8.5 πŸ•π’ƒ + πŸ’π’„; 𝒃 = πŸ“πŸŽ, 𝒄 = 𝟏𝟐𝟎 67
830 Yen 12 4 𝒙 = π’Ž + 𝒏 + 𝒅 + 𝒆 + 𝒇 = (-3) πŸ’ πŸ— + (-2) + (-27) + (-2.5) = -32 – πŸ— πŸπŸ– + πŸ– πŸπŸ– = -32 𝟏 πŸπŸ– 8

5 Expanding brackets To substitute numbers into expressions with brackets, it can be easier to expand them first. This lesson, we will expand each expression before substituting.

6 = 32 + 24 = 56 = 4(2π‘₯ + 3𝑦) 8π‘₯ + 12𝑦 Expanding brackets Example
Simplify 4(2π‘₯ + 3𝑦) and substitute π‘₯ = 4 and 𝑦 = 2 into it. 4(2π‘₯ + 3𝑦) = 8π‘₯ + 12𝑦 = That’s about it, try the last worksheets! 56 =

7 Answers 18 36 6 πŸπŸπ’‚π’ƒπŸπ’„ 72 πŸ‘(πŸ‘π’„ + πŸπ’ƒ) 1 πŸ“πŸ•πŸŽπŸŽ 𝒀𝒆𝒏 πŸ’(πŸ‘π’™ βˆ’ π’šπŸ) 510
πŸ‘π’‚+πŸ‘π’ƒ 18 πŸ–π’‚ – πŸ’π’ƒ 36 πŸ‘π’‚π’ƒ – π’‚π’ƒπŸ 6 πŸπŸπ’‚π’ƒπŸπ’„ 72 πŸ‘(πŸ‘π’„ + πŸπ’ƒ) πŸ‘π’™πŸ πŸ’π’š 1 πŸ‘ 𝟐 πŸ‘π’™πŸ πŸ’π’™ + πŸ–π’š πŸ“πŸ•πŸŽπŸŽ 𝒀𝒆𝒏 πŸ’(πŸ‘π’™ βˆ’ π’šπŸ) 510 πŸπŸπ’™ βˆ’ πŸ’π’šπŸ ( ) 8 -24 π’Š=𝟏 πŸ– 𝒙 π’Š

8 One last example. Example For π‘₯ + 𝑦 2 , simplify and substitute for π‘₯ = 1 and 𝑦 = 2. Let’s try with substitution first. (1 + 2)2 = 9 How about expanding first? π‘₯2 + 𝑦2? Mmm… = 5 Well that didn’t work. π‘₯ + 𝑦 2 β‰  π‘₯2 + 𝑦2. So what is it?

9 π‘₯2 π‘₯𝑦 𝑦π‘₯ 𝑦2 Let’s look at (x + y) 2 as an area. π‘₯ 𝑦 π‘₯
Here we want the total area. I made the size of π‘₯ look different to 𝑦 but this doesn’t matter. 𝑦π‘₯ 𝑦 𝑦2

10 So if we add each part together we get…
π‘₯ + 𝑦 2 = π‘₯2 + π‘₯𝑦 +𝑦π‘₯ + 𝑦2 = π‘₯2 + 2π‘₯𝑦 + 𝑦2 How about π‘₯ – 𝑦 2 ? And how about (π‘₯ + 𝑦)(π‘₯ – 𝑦)?


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