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How do you solve the system $y - 2x = 3$ , $x² + y² = 18$ ?

1 Answer

Nov 16, 2015

$(x,y) = (3/5, 4 1/5)$
or
$(x,y)=(-3,-3)$

Explanation:

Given
[1] $color(white)("XXX")y-2x=3$
[2] $color(white)("XXX")x^2+y^2 =18$

Re-writing [1]
[3] $color(white)("XXX")y=2x+3$

Substituting $(2x+3)$ from [3] for $y$ in [2]
[4] $color(white)("XXX")x^2+(2x+3)^2=18$

Simplifying
[5] $color(white)("XXX")x^2+4x^2+12x+9=18$

[6] $color(white)("XXX")5x^2+12x-9=0$

Factoring
[7] $color(white)("XXX")(5x-3)(x+3)=0$

$rArr x= 3/5$ or $x=-3$

case 1: $color(black)(x=3/5)$
Substituting $3/5$ for $x$ in [3]
$color(white)("XXX")y=2(3/5)+3= 4 1/5$

case 2: $color(black)(x=-3)$
Substituting $(-3)$ for $x$ in [3]
$color(white)("XXX")y = 2(-3) +3 = -3$

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