This another section where all the questions are in the B to A* range. I think that algebra’s reputation as being very difficult is exaggerated. Like all maths topics if you really get to grips with the basics, you’ll be able to tackle the more challenging stuff. If you’re struggling step back to a lower level and when you’re more confident come back to the higher level. A little repetition and practice goes a long way. You might not get it the first time but if you keep coming back to it, you’re more likely to succeed.

Further Simultaneous Equations

First some key facts:

When 2 straight lines cross or intersect they only do so in one point.

When you plot a quadratic and linear function on the same graph, there are potential outcomes:-

- The linear graph crosses or intersects the quadratic graph at 2 points
- The linear graph touches the quadratic graph at one point
- The linear graph and the quadratic graph do not intersect

Here’s the graph of **y = x² — 4x + 1 **and **y = x + 1**

The lines cross or intersect at 2 points, (0, 1) and (5, 6)

So you can solve these simultaneous equations by plotting the graphs but you also use algebra:-

1. Start with the linear equation and, if you need to, make x or y the subject.

So we have:-

**y = x + 1**

and

**y = x² — 4x + 1**

2. Next **substitute **the expression for “y” from the linear equation into the quadratic equation:-

**x + 1 = x ² — 4x + 1**

3. Adjust so that you move all the terms to the left hand side:-

**x + 1 = x ² — 4x + 1**

**x² — 4x + 1 = x + 1**

**x ² — 5x = 0**

**x(x — 5) = 0**

**So x = 0 or 5 **

*i.e. when x is 0 or 5, then x(x — 5) does equal zero.*** **

to be continued…