How do you solve the simultaneous equations $x²+y²=5$ and $y=3x+1$ ?

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Answer 1 · 2018-04-01T01:04:58

$x = 2/5,quad y=11/5$

or

$x = -1, y=2$

Substitute $y=3x+1$ in $x^2+y^2=5$ to obtain

$x^2+(3x+1)^2=5qquad implies qquad 10x^2+6x-4=0$

Hence

$5x^2+3x-2 = 0 qquad implies qquad 5x^2+5x-2x-2 = 0qquad implies$
$5x(x+1)-2(x+1) = 0 qquad implies qquad (5x-2)(x+1) = 0$

Hence

How do you solve the simultaneous equations x²+y²=5x²+y²=5 and y=3x+1 y=3x+1 ?