# Simultaneous Equations

Example –

Solve the simultaneous equations

$x^2+y^2=10$

and

$x+2y=5$.

This example requires solution via substitution, i.e. make either x or y the subject of one equation and insert it into the other. The obvious choice would be to make x the subject of the second equation – it is the quickest, least complicated choice. The second equation tells us that $x=5-2y$. We can insert this into the first equation: $(5-2y)^2+y^2=10$. By multiplying out the brackets and simplifying we see that this is a quadratic equation in y:

$(5-2y)^2+y^2=10$
$\Longrightarrow (5-2y)(5-2y)+y^2=10$
$\Longrightarrow 25-10y-10y+4y^2+y^2=10$
$\Longrightarrow 5y^2-20y+15=0$
$\Longrightarrow y^2-4y+3=0$
$\Longrightarrow (y-3)(y-1)=0$
This tells us that y has to be either 3 or 1. If $y=3$, then $x=5-2\times 3=-1$ (from the second equation) and if $y=1$ then $x=5-2\times 1=3$.

We obtain the solutions $(x_1,y_1)=(-1,3)$ and $(x_2,y_2)=(3,1)$.

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