How do you solve the system $y^{2} = x^{2} - 9$ $y^2=x^2-9$ and $2 y = x - 3$ $2y=x-3$ ?

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Answer 1 · 2016-04-22T17:54:11

${x = \pm 5, y = 4}$ ${x=±5,y=4}$

$y^{2} = x^{2} - 9$ $y^2=x^2-9$
$y^{2} + 9 = x^{2} (1)$ $color(green)(y^2+9)=x^2" "(1)$

$2 y = x - 3$ $2y=x-3$
$2 y + 3 = x$ $2y+3=x$
${(2 y + 3)}^{2} = x^{2} (2)$ $color(green)((2y+3)^2)=x^2" "(2)$

$y^{2} + 9 = {(2 y + 3)}^{2}$ $color(green)(y^2+9=color(green)((2y+3)^2))$

$y^{2} + 9 = 4 y^{2} + 12 y + 9$ $y^2+9=4y^2+12y+9$

$4 y^{2} - y^{2} + 12 y + 9 - 9 = 0$ $4y^2-y^2+12y+9-9=0$

$3 y^{2} + 12 y = 0$ $3y^2+12y=0$

$cancel(3)cancel(y)^2=cancel(12)cancel(y)$

$y=4$

$"using (1)"$

$4^2+9=x^2$

$16+9=x^2$

$x^2=25$

$x=±5$

${x=±5,y=4}$

How do you solve the system y^2=x^2-9y2=x2−9y^2=x^2-9 and 2y=x-32y=x−32y=x-3?