How do you solve the system $x^{2} + y^{2} = 20$ $x^2+y^2=20$ and $y = x - 4$ $y=x-4$ ?

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How do you solve the system $x^{2} + y^{2} = 20$ $x^2+y^2=20$ and $y = x - 4$ $y=x-4$ ?

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Answer 1 · 2016-11-22T14:08:19

$x = 2 \pm \sqrt{6}$ $x=2+-sqrt6$

Explanation:

Since we are given y in terms of x, we can substitute directly into the other equation.

$\Rightarrow x^{2} + {(x - 4)}^{2} = 20 \leftarrow equation in one variable$ $rArrx^2+(x-4)^2=20larr" equation in one variable"$

$\Rightarrow x^{2} + x^{2} - 8 x + 16 - 20 = 0 \leftarrow quadratic equation$ $rArrx^2+x^2-8x+16-20=0larr" quadratic equation"$

$\Rightarrow 2 x^{2} - 8 x - 4 = 0$ $rArr2x^2-8x-4=0$

divide both sides by 2

$x^{2} - 4 x - 2 = 0$ $x^2-4x-2=0$

To solve use the $quadratic formula$ $color(blue)"quadratic formula"$

with $a = 1, b = - 4, c = - 2$ $a=1,b=-4,c=-2$

$\Rightarrow x = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2 \sqrt{6}}{2} = 2 \pm \sqrt{6}$ $rArrx=(4+-sqrt24)/2=(4+-2sqrt6)/2=2+-sqrt6$

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How do you solve the system $x^{2} + y^{2} = 20$ $x^2+y^2=20$ and $y = x - 4$ $y=x-4$ ?

1 Answer

Explanation:

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How do you solve the system x^2+y^2=20x2+y2=20x^2+y^2=20 and y=x-4y=x−4y=x-4?

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How do you solve the system $x^{2} + y^{2} = 20$ $x^2+y^2=20$ and $y = x - 4$ $y=x-4$ ?