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© 2010 Pearson Prentice Hall. All rights reserved.
What you should learn
• Appreciate basic algebraic thinking
• Simplify algebraic expressions by using the rules of indices
1
© 2010 Pearson Prentice Hall. All rights reserved.
Review
Evaluate the following expressions given the values:
a) 𝑥 + 7 𝑥 = −6
b) 2𝑦 − 8 𝑦 = 3
c)
𝑎2
3
− 5𝑎 𝑎 = 6
d) 𝑥3
−
6𝑦
𝑥
𝑥 = −3, 𝑦 = 10
Simplify the Following:
a) 3𝑥 + 6𝑥
b) 10𝑥 − 8𝑦 + 7𝑥 − 6𝑐 + 𝑦 + 𝑐
c) 3𝑥2 − 6𝑥 + 5 − 8 + 𝑥 − 4𝑥2
2
© 2010 Pearson Prentice Hall. All rights reserved.
Review
Evaluate the following expressions given the values:
a) 𝑥 + 7 𝑥 = −6
−6 + 7 = 1
b) 2𝑦 − 8 𝑦 = 3
2 3 − 8 = 6 − 8 = −2
c)
𝑎2
3
− 5𝑎 𝑎 = 6
62
3
− 5 6 =
36
3
− 30 = 12 − 30 = −18
d) 𝑥3 −
6𝑦
𝑥
𝑥 = −3, 𝑦 = 10
(−3)3−
6 10
−3
= −27 −
60
−3
= −27 + 20 = −7 3
© 2010 Pearson Prentice Hall. All rights reserved.
Review
Evaluate the following expressions given the values:
a) 3𝑥 + 6𝑥 = 9𝑥
b) 10𝑥 − 8𝑦 + 7𝑥 − 6𝑐 + 𝑦 + 𝑐
= 10𝑥 + 7𝑥 − 8𝑦 + 𝑦 − 6𝑐 + 𝑐
= 17𝑥 − 7𝑦 − 5𝑐
c) 3𝑥2 − 6𝑥 + 5 − 8 + 𝑥 − 4𝑥2
= 3𝑥2 − 4𝑥2 − 6𝑥 + 𝑥 + 5 − 8
= −𝑥2 − 5𝑥 − 3
4
© 2010 Pearson Prentice Hall. All rights reserved.
Powers or Indices
A number written in this format has two parts 𝑥4
The Base is the number being multiplied
The Index/Power tells us how many times we multiply the base by itself
So:
𝑥4
= 𝑥 × 𝑥 × 𝑥 × 𝑥
Base
Index/Power
© 2010 Pearson Prentice Hall. All rights reserved.
Rules for indices
1. 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛: 𝑥𝑎 ∙ 𝑥𝑏 = 𝑥𝑎+𝑏 𝐼𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑤𝑒 𝑎𝑑𝑑 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠
2. Division:
𝑥𝑎
𝑥𝑏 = 𝑥𝑎−𝑏𝐼𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑤𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠
3. 𝑥0 = 1 𝐴𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜 𝑝𝑜𝑤𝑒𝑟 𝑒𝑞𝑢𝑎𝑙𝑠 1.
4. 𝑥−𝑎 =
1
𝑥𝑎 , 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑡𝑤𝑜 𝑟𝑢𝑙𝑒𝑠
5. 𝑥1
= 𝑥 𝐴 𝑛𝑢𝑚𝑏𝑒𝑟 𝑟𝑎𝑖𝑠𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑝𝑜𝑤𝑒𝑟 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑖𝑡𝑠𝑒𝑙𝑓.
6. (𝑥𝑎)𝑏 = 𝑥𝑎𝑏 𝑊𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 𝑤ℎ𝑒𝑛 𝑜𝑛𝑒 𝑖𝑠 𝑟𝑎𝑖𝑠𝑒𝑑 𝑡𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟
© 2010 Pearson Prentice Hall. All rights reserved.
When simplifying expressions by multiplying or dividing we combine the
terms(They don’t have to be like terms) and simplify them using the rules
of indices.
7
© 2010 Pearson Prentice Hall. All rights reserved.
Multiplication Examples
1. 3𝑥 × 4𝑥2
= 3 ∙ 4𝑥1+2
= 12𝑥3
2. 7𝑎3𝑏7 × 12𝑏2
= 7 ∙ 12𝑎3𝑏7+2
= 84𝑎3
𝑏9
8
© 2010 Pearson Prentice Hall. All rights reserved.
Multiplication Examples
3. 0.45𝑥5𝑦−3𝑧 ∙ 8𝑥2𝑦10𝑧2
= 0.45 ∙ 8𝑥5+2𝑦−3+10𝑧1+2
= 3.6𝑥7𝑦7𝑧3
4. −5𝑝7
𝑞5
× −3𝑝5
𝑞−7
× 2𝑝3
𝑞2
= −5 −3 2 𝑝7+5+3
𝑞5−7+2
= 30𝑝15𝑞0
= 30𝑝15
9
© 2010 Pearson Prentice Hall. All rights reserved.
Try
1. 5𝑥7𝑦3 ∙ 10𝑥2𝑦−2
2. −4𝑎3
𝑏−5
𝑐 × 7𝑎6
𝑏2
𝑐−1
10
= 5 ∙ 10𝑥7+2
𝑦3−2
= 50𝑥9
𝑦
= −4 ∙ 7𝑎3+6𝑏−5+2𝑐1−1
= −28𝑎9𝑏−3𝑐0
=
−28𝑎9
𝑏3
© 2010 Pearson Prentice Hall. All rights reserved.
Division Examples
1. 2𝑥3 ÷ 4𝑥2
=
2
4
𝑥3−2
=
𝑥
2
or 0.5𝑥
2.
12𝑎3𝑏7𝑐
3𝑎4𝑏3𝑐5
= 4𝑎3−4
𝑏7−3
𝑐1−5
= 4𝑎−1
𝑏4
𝑐−4
=
4𝑏4
𝑎𝑐4
11
© 2010 Pearson Prentice Hall. All rights reserved.
Division Examples
3.
7𝑥𝑦8𝑧3
5𝑥2𝑦6𝑧2
=
7𝑥1−2
𝑦8−6
𝑧3−2
5
=
7𝑥−1𝑦2𝑧
5
=
7𝑦2𝑧
5𝑥 12
© 2010 Pearson Prentice Hall. All rights reserved.
Try
1. 10𝑦3 ÷ 5𝑦6
2.
24𝑎5𝑏3𝑐2
16𝑎3𝑏5𝑐10
13
=
3𝑎5−3𝑏3−5𝑐2−10
2
=
3𝑎2𝑏−2𝑐−8
2
=
3𝑎2
2𝑏2𝑐8
= 2𝑦3−6
= 2𝑦−3
=
2
𝑦3
© 2010 Pearson Prentice Hall. All rights reserved.
Try
1.
3𝑥6𝑦∙4𝑥3𝑦7
8𝑥3𝑦10
2.
6𝑎3𝑏5𝑐4×3𝑎3𝑏5
3𝑎5𝑏2×10𝑎3𝑐5
14
=
3 ∙ 4𝑥6+3−3
𝑦1+7−10
8
=
12𝑥6𝑦−2
8
=
3𝑥6
2𝑦2
=
6 ∙ 3𝑎3+3𝑏5+5𝑐4
3 ∙ 10𝑎5+3𝑏2𝑐5
=
18𝑎6−8𝑏10−2𝑐4−5
30
=
3𝑎−2𝑏8𝑐−1
5
=
3𝑏8
5𝑎2𝑐

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Multiplication and division of algebraic terms.pptx

  • 1. © 2010 Pearson Prentice Hall. All rights reserved. What you should learn • Appreciate basic algebraic thinking • Simplify algebraic expressions by using the rules of indices 1
  • 2. © 2010 Pearson Prentice Hall. All rights reserved. Review Evaluate the following expressions given the values: a) 𝑥 + 7 𝑥 = −6 b) 2𝑦 − 8 𝑦 = 3 c) 𝑎2 3 − 5𝑎 𝑎 = 6 d) 𝑥3 − 6𝑦 𝑥 𝑥 = −3, 𝑦 = 10 Simplify the Following: a) 3𝑥 + 6𝑥 b) 10𝑥 − 8𝑦 + 7𝑥 − 6𝑐 + 𝑦 + 𝑐 c) 3𝑥2 − 6𝑥 + 5 − 8 + 𝑥 − 4𝑥2 2
  • 3. © 2010 Pearson Prentice Hall. All rights reserved. Review Evaluate the following expressions given the values: a) 𝑥 + 7 𝑥 = −6 −6 + 7 = 1 b) 2𝑦 − 8 𝑦 = 3 2 3 − 8 = 6 − 8 = −2 c) 𝑎2 3 − 5𝑎 𝑎 = 6 62 3 − 5 6 = 36 3 − 30 = 12 − 30 = −18 d) 𝑥3 − 6𝑦 𝑥 𝑥 = −3, 𝑦 = 10 (−3)3− 6 10 −3 = −27 − 60 −3 = −27 + 20 = −7 3
  • 4. © 2010 Pearson Prentice Hall. All rights reserved. Review Evaluate the following expressions given the values: a) 3𝑥 + 6𝑥 = 9𝑥 b) 10𝑥 − 8𝑦 + 7𝑥 − 6𝑐 + 𝑦 + 𝑐 = 10𝑥 + 7𝑥 − 8𝑦 + 𝑦 − 6𝑐 + 𝑐 = 17𝑥 − 7𝑦 − 5𝑐 c) 3𝑥2 − 6𝑥 + 5 − 8 + 𝑥 − 4𝑥2 = 3𝑥2 − 4𝑥2 − 6𝑥 + 𝑥 + 5 − 8 = −𝑥2 − 5𝑥 − 3 4
  • 5. © 2010 Pearson Prentice Hall. All rights reserved. Powers or Indices A number written in this format has two parts 𝑥4 The Base is the number being multiplied The Index/Power tells us how many times we multiply the base by itself So: 𝑥4 = 𝑥 × 𝑥 × 𝑥 × 𝑥 Base Index/Power
  • 6. © 2010 Pearson Prentice Hall. All rights reserved. Rules for indices 1. 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛: 𝑥𝑎 ∙ 𝑥𝑏 = 𝑥𝑎+𝑏 𝐼𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑤𝑒 𝑎𝑑𝑑 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 2. Division: 𝑥𝑎 𝑥𝑏 = 𝑥𝑎−𝑏𝐼𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑤𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 3. 𝑥0 = 1 𝐴𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜 𝑝𝑜𝑤𝑒𝑟 𝑒𝑞𝑢𝑎𝑙𝑠 1. 4. 𝑥−𝑎 = 1 𝑥𝑎 , 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑡𝑤𝑜 𝑟𝑢𝑙𝑒𝑠 5. 𝑥1 = 𝑥 𝐴 𝑛𝑢𝑚𝑏𝑒𝑟 𝑟𝑎𝑖𝑠𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑝𝑜𝑤𝑒𝑟 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑖𝑡𝑠𝑒𝑙𝑓. 6. (𝑥𝑎)𝑏 = 𝑥𝑎𝑏 𝑊𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 𝑤ℎ𝑒𝑛 𝑜𝑛𝑒 𝑖𝑠 𝑟𝑎𝑖𝑠𝑒𝑑 𝑡𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟
  • 7. © 2010 Pearson Prentice Hall. All rights reserved. When simplifying expressions by multiplying or dividing we combine the terms(They don’t have to be like terms) and simplify them using the rules of indices. 7
  • 8. © 2010 Pearson Prentice Hall. All rights reserved. Multiplication Examples 1. 3𝑥 × 4𝑥2 = 3 ∙ 4𝑥1+2 = 12𝑥3 2. 7𝑎3𝑏7 × 12𝑏2 = 7 ∙ 12𝑎3𝑏7+2 = 84𝑎3 𝑏9 8
  • 9. © 2010 Pearson Prentice Hall. All rights reserved. Multiplication Examples 3. 0.45𝑥5𝑦−3𝑧 ∙ 8𝑥2𝑦10𝑧2 = 0.45 ∙ 8𝑥5+2𝑦−3+10𝑧1+2 = 3.6𝑥7𝑦7𝑧3 4. −5𝑝7 𝑞5 × −3𝑝5 𝑞−7 × 2𝑝3 𝑞2 = −5 −3 2 𝑝7+5+3 𝑞5−7+2 = 30𝑝15𝑞0 = 30𝑝15 9
  • 10. © 2010 Pearson Prentice Hall. All rights reserved. Try 1. 5𝑥7𝑦3 ∙ 10𝑥2𝑦−2 2. −4𝑎3 𝑏−5 𝑐 × 7𝑎6 𝑏2 𝑐−1 10 = 5 ∙ 10𝑥7+2 𝑦3−2 = 50𝑥9 𝑦 = −4 ∙ 7𝑎3+6𝑏−5+2𝑐1−1 = −28𝑎9𝑏−3𝑐0 = −28𝑎9 𝑏3
  • 11. © 2010 Pearson Prentice Hall. All rights reserved. Division Examples 1. 2𝑥3 ÷ 4𝑥2 = 2 4 𝑥3−2 = 𝑥 2 or 0.5𝑥 2. 12𝑎3𝑏7𝑐 3𝑎4𝑏3𝑐5 = 4𝑎3−4 𝑏7−3 𝑐1−5 = 4𝑎−1 𝑏4 𝑐−4 = 4𝑏4 𝑎𝑐4 11
  • 12. © 2010 Pearson Prentice Hall. All rights reserved. Division Examples 3. 7𝑥𝑦8𝑧3 5𝑥2𝑦6𝑧2 = 7𝑥1−2 𝑦8−6 𝑧3−2 5 = 7𝑥−1𝑦2𝑧 5 = 7𝑦2𝑧 5𝑥 12
  • 13. © 2010 Pearson Prentice Hall. All rights reserved. Try 1. 10𝑦3 ÷ 5𝑦6 2. 24𝑎5𝑏3𝑐2 16𝑎3𝑏5𝑐10 13 = 3𝑎5−3𝑏3−5𝑐2−10 2 = 3𝑎2𝑏−2𝑐−8 2 = 3𝑎2 2𝑏2𝑐8 = 2𝑦3−6 = 2𝑦−3 = 2 𝑦3
  • 14. © 2010 Pearson Prentice Hall. All rights reserved. Try 1. 3𝑥6𝑦∙4𝑥3𝑦7 8𝑥3𝑦10 2. 6𝑎3𝑏5𝑐4×3𝑎3𝑏5 3𝑎5𝑏2×10𝑎3𝑐5 14 = 3 ∙ 4𝑥6+3−3 𝑦1+7−10 8 = 12𝑥6𝑦−2 8 = 3𝑥6 2𝑦2 = 6 ∙ 3𝑎3+3𝑏5+5𝑐4 3 ∙ 10𝑎5+3𝑏2𝑐5 = 18𝑎6−8𝑏10−2𝑐4−5 30 = 3𝑎−2𝑏8𝑐−1 5 = 3𝑏8 5𝑎2𝑐