Further Algebra

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

Graph of quadratic function y = x² -4x + 1 and linear function 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


y = x² — 4x + 1

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

x + 1 = x² — 4x + 1

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

x + 1 = x² — 4x + 1

x² — 4x + 1 = x + 1

x² — 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…

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