We are moving on to a new topic now in algebra and it builds on much of what you have already learned in grades 8 and 9 so here is some revision:
An identity is a mathematical statement that shows how an expression can be stated in a different form e.g.
3(a+1)= 3a + 3.
Are the following identities?
(a) 3b²; (b) 16b² + 4ab = 4b(4b + a); (c) 16b² + 4ab ÷4b = 4b + a; (d) 4a(b +3) = 4ab + 7
In the case of (a) it is a single term and is therefore not an identity.
In the case of (d) the statement is not true (it is a false statement) so it is also not an identity.
An equation is a statement which says that something (an expression) is equal to / the same as something else e.g. 9 = 3²Equations can be fairly simple like the example above or very complex. To start with we will only deal with simple equations called linear equations. Linear equations are used very often in everyday life when we solve simple problems for which we have some information but one piece of information is missing e.g.
Pies cost R17.00 each and I wish to buy for four people. How much money do I need? This can be written as:
4 X R17.00 = a We easily work out that the answer is R68.00
We could have an even more complicated problem: I have R90.00 and I want to buy pies that cost R17.00 each for five people, Can I afford it and will I have any change? This can be stated as:
90 -5(17) = x This gives us the answer 5 which is positive so our answer is change of R5, but if I wanted to buy 6 pies the equation would be 90 – 6(17) = x and the answer would be -12 showing that I could not afford the pies as I would be short of R12.00 because the answer is negative.
Sometimes linear equations help us to see the relationship between two quantities e.g. ax = b tells us that we can also consider x to be the quotient of b and a because x =b/a or to use the example of the pies again a(Number of pies) x b(price of pies) = c(total cost)
ab = c If I spent R60.00 on 4 pies (c/a) then the price (b) would be easily calculated c/a = b 60/4 = b = R15.00
Notice what we did to solve the equation for an unknown term (b):
- We write out the given equation: ab = c
2. Then we simplify both sides by removing brackets and multiplying by LCD to remove fractions
3. Then we move all the known terms to the right of the = sign by performing the same operation on both sides e.g. to remove a from ab I divide by a on the left so now I divide the right by a as well => b = c/a
4. Then, if we are able to, we substitute the values of the know terms b = 60/4 = 15
So here are a few simple linear equations you can solve for x:
x + c = d
-3x =-6
7x +8 = 12x -7
5(x -12) – (x-40) = 12
3x – 3(x-4) =12
ex – f =g
h(x +k) =m(x +1)
Coming up next: Quadratic equations!