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# 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 $y = 3x$ and $y = x^2 + 2$ simultaneously.

Substitute $y=3x$ into $y=x^2+2$

$3x=x^2 +2$

Rearrange $3x=x^2 +2$ to equal 0

$x^2-3x+2=0$

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

The correct factorisation is $(x-1)(x-2)$

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

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

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

Substitute $x=2$ into $y=3x$

$y=3\times2$ so $y=6$

That's it!

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

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

Remember which $x$ goes with which $y$!

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

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

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