basic algebra the use of letters to replace a word and/or an unknown value
TRANSCRIPT
![Page 1: Basic Algebra The use of letters to replace a word and/or an unknown value](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/1.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/2.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/3.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/4.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/5.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/6.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/7.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/8.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/9.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/10.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/11.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/12.jpg)
•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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/13.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/14.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/15.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/16.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/17.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/18.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/19.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/20.jpg)
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](https://reader034.vdocument.in/reader034/viewer/2022051416/56649e405503460f94b31195/html5/thumbnails/21.jpg)
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