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Year 7 Brackets Dr J Frost

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Presentation on theme: "Year 7 Brackets Dr J Frost "β€” Presentation transcript:

1 Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
Year 7 Brackets Dr J Frost Objectives: Be able to expand and simplify expressions involving single brackets. Last modified: 1st September 2015

2 Brackets If I want β€œ3 lots of 2π‘₯+4”, what will I have? πŸ”π’™+𝟏𝟐
What’s another way I can write β€œ3 lots of 2π‘₯+4”? πŸ‘ πŸπ’™+πŸ’ ? ? So we can multiply each thing inside the bracket by the thing outside it. 2 π‘₯ We can also see this using areas. Area of whole rectangle =πŸ’(𝟐+𝒙) Total area of small rectangles =πŸ–+πŸ’π’™ 4 ? ?

3 2 5+π‘₯ =𝟏𝟎+πŸπ’™ 7 2π‘₯βˆ’3𝑦 =πŸπŸ’π’™βˆ’πŸπŸπ’š 2π‘₯ 3π‘₯+5π‘₯𝑦 =πŸ” 𝒙 𝟐 +𝟏𝟎 𝒙 𝟐 π’š
Examples 2 5+π‘₯ =𝟏𝟎+πŸπ’™ 7 2π‘₯βˆ’3𝑦 =πŸπŸ’π’™βˆ’πŸπŸπ’š 2π‘₯ 3π‘₯+5π‘₯𝑦 =πŸ” 𝒙 𝟐 +𝟏𝟎 𝒙 𝟐 π’š 8π‘₯𝑦 1βˆ’3π‘₯𝑦 =πŸ–π’™π’šβˆ’πŸπŸ’ 𝒙 𝟐 π’š 𝟐 1 2 π‘₯ π‘₯+4 = 𝟏 𝟐 𝒙 𝟐 +πŸπ’™ ? ? ? ? ?

4 βˆ’ (π‘₯βˆ’3) 1 =βˆ’1π‘₯+3 =βˆ’π‘₯+3 Dealing with negative sign ? ?
Click to give hint > βˆ’ (π‘₯βˆ’3) 1 =βˆ’1π‘₯+3 ? =βˆ’π‘₯+3 ?

5 Expanding and Simplifying
If we have multiple brackets we can usually collect like terms after. 1+2 3+π‘₯ =𝟏+πŸ”+πŸπ’™ =πŸ•+πŸπ’™ 7βˆ’2 3βˆ’2π‘₯ =πŸ•βˆ’πŸ”+πŸ’π’™ =𝟏+πŸ’π’™ π‘Ž π‘Žβˆ’π‘ βˆ’π‘ π‘βˆ’π‘Ž = 𝒂 𝟐 βˆ’π’‚π’ƒβˆ’ 𝒃 𝟐 +𝒂𝒃 = 𝒂 𝟐 βˆ’ 𝒃 𝟐 3π‘₯ 3π‘₯βˆ’4 βˆ’ 5βˆ’2π‘₯ =πŸ— 𝒙 𝟐 βˆ’πŸπŸπ’™βˆ’πŸ“+πŸπ’™ =πŸ— 𝒙 𝟐 βˆ’πŸπŸŽπ’™βˆ’πŸ“ ? ? ? ?

6 Test Your Understanding
Expand and simplify. 11 4 π‘₯ 2 +3π‘¦βˆ’2 =πŸ’πŸ’ 𝒙 𝟐 +πŸ‘πŸ‘π’šβˆ’πŸπŸ 5 3π‘₯βˆ’1 βˆ’4 2βˆ’π‘₯ =πŸπŸ“π’™βˆ’πŸ“βˆ’πŸ–+πŸ’π’™ =πŸπŸ—π’™βˆ’πŸπŸ‘ 5βˆ’4 2βˆ’π‘₯ =πŸ“βˆ’πŸ–+πŸ’π’™ =πŸ’π’™βˆ’πŸ‘ ? ? ? Bro Note: βˆ’3+4π‘₯ is also acceptable, but 4π‘₯βˆ’3 is preferable. Can you think why?

7 Exercises 1 Expand (and where relevant simplify) the following: 2 π‘₯+4 =πŸπ’™+πŸ– 9 π‘₯βˆ’3 =πŸ—π’™βˆ’πŸπŸ• 3 2π‘₯+1 =πŸ”π’™+πŸ‘ 9 9βˆ’2π‘₯ =πŸ–πŸβˆ’πŸπŸ–π’™ 2 π‘₯+1 +3 π‘₯+2 =πŸ“π’™+πŸ– 7 2π‘₯+3 +2 π‘₯βˆ’4 =πŸπŸ”π’™+πŸπŸ‘ 5 3π‘₯βˆ’2 βˆ’7π‘₯=πŸ–π’™βˆ’πŸπŸŽ 7 2π‘₯βˆ’4 βˆ’3 π‘₯βˆ’2 =πŸπŸπ’™βˆ’πŸπŸ 2βˆ’ π‘₯βˆ’1 =πŸ‘βˆ’π’™ 3βˆ’ 1βˆ’π‘₯ =𝒙+𝟐 10βˆ’2 4+π‘₯ =πŸβˆ’πŸπ’™ βˆ’ 2βˆ’3π‘₯ =πŸ‘π’™βˆ’πŸ 5βˆ’3 1βˆ’4π‘₯ =πŸπŸπ’™+𝟐 Expand and simplify the following: 2 π‘₯ 2 3π‘¦βˆ’4 =πŸ” 𝒙 𝟐 π’šβˆ’πŸ– 𝒙 𝟐 5π‘₯𝑦 3π‘₯+2𝑦 =πŸπŸ“ 𝒙 𝟐 π’š+πŸπŸŽπ’™ π’š 𝟐 π‘₯𝑦 3+2π‘₯ βˆ’2π‘₯ 𝑦+2 =π’™π’š+𝟐 𝒙 𝟐 π’šβˆ’πŸ’π’™ π‘Ž π‘Žπ‘+π‘Ž βˆ’π‘ π‘Ž 2 +π‘Žπ‘ = 𝒂 𝟐 βˆ’π’‚ 𝒃 𝟐 3π‘₯ 𝑦 2 3π‘₯𝑦+4𝑦 =πŸ— 𝒙 𝟐 π’š πŸ‘ +πŸπŸπ’™ π’š πŸ‘ 10 π‘₯ 10 𝑦 π‘₯βˆ’ π‘₯ 10 𝑦 10 =𝟏𝟎𝟎 𝒙 𝟏𝟏 π’š 𝟏𝟎 βˆ’πŸπŸŽ 𝒙 𝟐𝟎 π’š 𝟏𝟎 5 4βˆ’3 2βˆ’ 1βˆ’π‘₯ =πŸ“βˆ’πŸπŸ“π’™ π‘₯ 1 π‘₯ + 1 π‘₯ 2 =𝟏+ 𝟏 𝒙 3 Find the area and perimeter of the following (expand and simplify your answer) a ? b ? 2π‘Ž π‘₯+3 c ? d ? 1 ? e 5 ? f 5 g ? h ? i ? 𝐴=πŸ“π’™+πŸπŸ“ 𝑃=πŸπ’™+πŸπŸ” ? 2 j ? ? k ? 𝐴=πŸπŸŽπ’‚βˆ’πŸ’ πŸπ’‚βˆ’πŸ =πŸπ’‚+πŸ– 𝑃=πŸπ’‚+𝟏𝟎 ? l ? m ? ? 2 [IMC 2004 Q22] In a maths exam with 𝑁 questions, you score π‘š marks for a correct answer to each of the first π‘ž questions and π‘š+2 marks for a correct answer to each of the remaining questions. What is the maximum possible score? A π‘š+2 π‘βˆ’2π‘ž B π‘π‘š C π‘šπ‘ž+ π‘š+2 π‘ž D 𝑁 π‘š+1 E π‘π‘š+π‘ž(π‘š+2) Solution: A 4 a ? b ? c ? d ? e ? f ? g ? ? h ?


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