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Non-linear Simultaneous Equations

Non-linear Simultaneous Equations

Non-linear Simultaneous Equations

We can find the points of intersection between straight (linear) lines and quadratic lines on a graph.

Sometimes you will be given two equations to solve: one linear and one non-linear (usually a quadratic). You can solve these by eliminating one of the variables from the quadratic and then solving that equation.

We can usually expect there to be 2 points that are common between a linear and a quadratic equation. This is because the quadratic is shaped like a U when drawn on a graph. This may not always be the case, however.

For example, we might be asked to solve y=3xy = 3x and y=x2+2y = x^2 + 2 simultaneously.


Substitute y=3xy=3x into y=x2+2y=x^2+2

3x=x2+23x=x^2 +2


Rearrange 3x=x2+23x=x^2 +2 to equal 0



Factorise x23x+2=0x^2-3x+2=0


The correct factorisation is (x1)(x2)(x-1)(x-2)

21=3-2-1=-3, and 2×1=2-2\times -1=2. Therefore, we can put these factors in brackets with xx. We can now treat these as two separate equations.


One solution is x=1x=1. Solve x2=0x-2=0


Substitute x=1x=1 into y=3xy=3x. What is the value of yy?


Substitute x=2x=2 into y=3xy=3x

y=3×2y=3\times2 so y=6y=6


That's it!

When x=1x=1,y=3y=3 and when x=2x=2,y=6y=6


This means that the lines would cross at (1,3)(1,3)and(2,6)(2,6)

Remember which xx goes with which yy!

What quadratic equation do you obtain if you eliminate y from: y=4xy = 4x and y=x2+3y = x^2 + 3? Give your answer in the form y=ax2+bx+cy=ax^2+bx+c

What two values of xx solve x24x+3=0 x^2 - 4x + 3=0?

As x=1x = 1 and x=3x = 3, what are the solutions to y=4xy = 4x and y=x2+3y = x^2 + 3?