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

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 top-grades! (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 x-coordinate is 2.25 and the y-coordinate is 0.5 Create a common coefficient      (3)     (1)×2    (4)     (2)×3   (3)+(4) See if you can show that:

 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 short-cut.

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 short-cut.

So the coefficient is -808

The general case:

To find the term of  in the expansion of   you would use the following short-cut.

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