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Basic Algebra The use of letters to replace a word and/or an unknown value

Basic Algebra The use of letters to replace a word and/or an unknown value

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Page 1: Basic Algebra The use of letters to replace a word and/or an unknown value

Basic Algebra

The use of letters to replace a word and/or

an unknown value

Page 2: Basic Algebra The use of letters to replace a word and/or an unknown value

10 Mars bars cost £2.50 or 250 pence, can be rewritten as:

10m = 250 (pence)

This means EXACTLY the same thing as the sentence but can be easier to read, especially if the statement is a lot longer (next slide)!!

Page 3: Basic Algebra The use of letters to replace a word and/or an unknown value

E.g. 6 eggs, 10 cakes, 5 apples and 4 pears cost a total of £4.25

Can be written as:

6e + 10c + 5a + 4p = 4.25The letter stands for an object (eggs, cakes, etc.) and the letter has a value (or individual cost).

Page 4: Basic Algebra The use of letters to replace a word and/or an unknown value

If we look again at the first statement (now called an EQUATION because of the equal sign)

10m = 250

From this we can work out the cost of m (one Mars bar)

Page 5: Basic Algebra The use of letters to replace a word and/or an unknown value

If 10m = 250

Then m (we do not need to put in the “1”, more on this later) is simply:

m = 250 ÷ 10 = 25

Since part of the original statement gave the cost in pence, then the answer must be given in the appropriate units (pence)!

m = 25p

Page 6: Basic Algebra The use of letters to replace a word and/or an unknown value

As long as you remember that the

letters in algebra stand for objects and that those objects have a

value, then things should be a bit easier!!

Page 7: Basic Algebra The use of letters to replace a word and/or an unknown value

Simplification of Terms

This is just getting the same ‘bits’ (or like terms) together…

Example: 5a + 3b – 2a + 6b

Remember “a” could be apples and “b” could be bananas. 5a means 5 × a but we drop the × sign as it could be confused with the letter “x”. Think if we didn’t drop the times sign how could we show 6x…..6×x?

Page 8: Basic Algebra The use of letters to replace a word and/or an unknown value

Also we show a single item “a” as simply the lone letter…we do not put a “1” in front. There is no need. ‘a’ indicates one apple (or 1a) so using a “1” is simply a waste of pen!!!

Back to the example 5a + 3b – 2a +6b

Get the like terms (or like ‘bits’) together and remember that the sign joins the letter that follows!!!!

5a – 2a + 3b + 6b

Do the maths, remember about negative numbers!!

3a + 9b

Page 9: Basic Algebra The use of letters to replace a word and/or an unknown value

If a letter has a power, such as x3, then treat it as a separate ‘object’ or

animal from similar terms such as 6x or 2x2 for example, UNLESS we are required to multiply or divide the terms!!!

Page 10: Basic Algebra The use of letters to replace a word and/or an unknown value

Now try these questions

1. 2x+3y-x+2y2. 3c+5d-c+6d+4c3. 4p-3q+2p+5q4. 5r-3s-7s+2r-s5. 5a-7b-4a-2b+12b6. 5ef+7gh-

9ef+2gh

x + 5y6c + 11d6p + 2q7r - 11sa + 3b- 4ef + 9gh

Page 11: Basic Algebra The use of letters to replace a word and/or an unknown value

Multiplying Expressions

Example:

2a × 4a becomes 8a2

Multiply the numbers then the letter!!

Just like 2 × 2 = 22

So a × a = a2

And so 2a x 4a must equal 8a2

Page 12: Basic Algebra The use of letters to replace a word and/or an unknown value

•If we multiply, for example, 3a2 × 2a3, we do the same things;

•That is the numbers first (3 × 2) then the powers (a2 × a3). Just remember the rules on multiplying powers. Which is……….?

•3a2 × 2a3 = 6a5

•If you can’t remember, write out the expression in its simplest terms.

Page 13: Basic Algebra The use of letters to replace a word and/or an unknown value

3a2 × 2a3

3 × a × a × 2 × a × a × a

Get the same ‘bits’ together

3 × 2 × a × a × a × a × a

This ‘bit’ means a5

Multiplying out gives:

6a5

Page 14: Basic Algebra The use of letters to replace a word and/or an unknown value

What about if there are different letters to be multiplied?

No problem; just do as the previous slide….

Example:

4a × 3b becomes 12 ab

4 × a × 3 × b

4 × 3 × a × b

12ab

Get the same ‘bits together, then do the maths

Page 15: Basic Algebra The use of letters to replace a word and/or an unknown value

A bit about convention on writing expressions in algebra.

1. Numbers in front of the letters

2. Letters in alphabetic order

3. High powers first

4. ‘Lone’ numbers last

Example: 3ax3 – 2tx2 + 5px -7

Page 16: Basic Algebra The use of letters to replace a word and/or an unknown value

Slightly harder algebra

Expanding brackets. 2(3a + 4b)REMEMBER everything inside is

multiplied by outside the brackets.

2(3a + 4b)This means… 2 × 3a + 2 × 4b

6a + 8b

Page 17: Basic Algebra The use of letters to replace a word and/or an unknown value

One thing you must remember is the sign of what is outside the bracket, in the previous example was not given, so it is assumed to be +

–2a(3a – 2b)

– 2a × 3a – 2a × – 2b

–6a2 + 4abRemember that the – sign between the 3a and the 2b relates to the 2b

Page 18: Basic Algebra The use of letters to replace a word and/or an unknown value

Sometimes we need to multiply out a pair of brackets. Remember the phrase FOIL.

F FirstO OuterI InnerL Last

(3x +3)(3x – 6)

Page 19: Basic Algebra The use of letters to replace a word and/or an unknown value

F gives 9x2 O gives – 18xI gives + 9xL gives – 18

Giving 9x2 – 18x + 9x – 18

Combining these terms gives

9x2 – 9x – 18This is a quadratic expression---more later!!

(3x +3)(3x – 6)

Page 20: Basic Algebra The use of letters to replace a word and/or an unknown value

There are several other methods, such as the ‘smiley face’.

(3x + 3)(3x – 6)

Whichever method you use you will get the same answer. Keep with it if you are happy and always get the right answer!!

Page 21: Basic Algebra The use of letters to replace a word and/or an unknown value

Now try these

1. 2(a+3)2. 3(2p+3)3. 7(4x-3)4. 9(3b+4c-2d)Expand AND simplify5. 4(4a+5)-4(5a+2b)6. (a+1)(a+2)7. (3e-5)(2e-1)8. (3p-1)(3p+1)

2a + 6 6p + 9 28x – 21 27b + 36c –

18d

– 4a – 8b +20 a2 +3a +2 6e2 – 13e + 5 9p2 – 1