How do you solve the system of equations $x^2+y^2=13$ and $2x-y=4$ ?

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Answer 1 · 2017-05-08T19:53:25

$x=3$

$y=2$

First, think of which two perfect squares would add up to $13$ :
$4+9=13$ or $9+4=13$

Take the square root of those perfect squares:
$sqrt4=2$ and $sqrt9=3$

$x$ must equal either $2$ or $3$ , and $y$ must equal $2$ or $3$

Let's say $x=2$ and $y=3$ :
$2(2)-(3)=4$ $rarr$ $4-3=4$ $rarr$ $1!=4$

Since that didn't work, let's say $x=3$ and $y=2$ :
$2(3)-(2)=4$ $rarr$ $6-2=4$ $rarr$ $4=4$

Therefore, $x=3$ and $y=2$

How do you solve the system of equations x^2+y^2=13x^2+y^2=13 and 2x-y=42x-y=4?