How do you solve the system $x^2 - x - y = 2$ , $4x - 3y = 0$ ?

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Answer 1 · 2015-12-21T20:22:11

$(-2/3,-8/9),(3,4)$

First modify the equation $x^2-x-y=2$ to solve for $y$ :

$y=x^2-x-2$

Now, take this and plug it into $4x-3y=0$

$4x-3(x^2-x-2)=0$

$4x-3x^2+3x+6=0$

$-3x^2+7x+6=0$

$3x^2-7x-6=0$

$(3x^2-9x)+(2x-6)=0$

$3x(x-3)+2(x-3)=0$

$(3x+2)(x-3)=0$

Thus, we have two possible values for $x$ : ${(x=-2/3),(x=3):}$

Plug these both into $4x-3y=0$

When $x=-2/3$ :

$4(-2/3)-3y=0$
$-8/3=3y$
$y=-8/9$

Solution point: $(-2/3,-8/9)$

When $x=3$ :

$4(3)-3y=0$

$12=3y$

$y=4$

Solution point: $(3,4)$

graph{(x^2-x-y-2)(4x-3y)=0 [-6.92, 13.08, -3.04, 6.96]}

How do you solve the system x^2 - x - y = 2x^2 - x - y = 2, 4x - 3y = 04x - 3y = 0?