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Core 1: Algebraic Skills Part 2
In this section, you learn some new
skills, in particular how to solve simultaneous equations.
You will apply these to
geometrical situations
1)
Linear simultaneous
equations
2)
Quadratic simultaneous
equations
These
skills will then be explored in more detail in the chapter on geometry
3)
Binomial expansion for
positive integer powers
4)
Understanding factorials
5)
Finding individual
coefficients from binomial expansion
6)
Harder example, for those
wishing to go for the topgrades! (Surely that is all of you...)
Linear simultaneous equations
Solve: (1)(2) Substitute this value into (2) 
Solving simultaneous equations finds where the graphs
intersect. We have therefore found that the red and blue line intersect
where the xcoordinate is 2.25 and the ycoordinate is 0.5 
Create a common coefficient (3)
(1)×2 (4)
(2)×3 (3)+(4) 
See if you can show that: 
Quadratic simultaneous equations
So 
These graphs intersect in two places, so there should be two
possible solutions. 
Because _{} 
You can see the graphs intersect at (2. 1) and (1, 0) 
Binomial expansion











_{} 


This might appear a little
daunting at first.
However, the coefficients follow a
very simple pattern which you may well know from Pascal’s triangle
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 To generate the next
line observe how each number is the sum of the two numbers above it 
Examples:




Understanding factorials and
combinations
Before we proceed to the next stage
we need to understand some new terminology
Factorials
Important: 0!=1
Combinations:
It is worth being aware that:
Finding coefficients:
Particular case:
1) Suppose that you wanted to find the
coefficient of in the expansion of
Rather than expand the whole thing you
can use the following shortcut.
So the coefficient is 560
2) Suppose that you wanted to find the
coefficient of in the expansion of
Again, rather than expand the whole thing
you can use the following shortcut.
So the coefficient is 808
The general case:
To find the term of in the expansion of you would use the
following shortcut.
Harder example
Find the term which is independent
of x (which means that one of the terms has no x involved!) in the expansion
of:
Can you see how to do this?
Well.....watch this....
We will work out the term:
As you can see, the x’s cancelled
out.
So the term independent of x is
1120