How do you solve y² = x + 3 and x - 2y = 12?

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How do you solve y² = x + 3 and x - 2y = 12?

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Answer 1 · 2018-06-10T12:39:42

If $y = 5$ $y=5$ , $x = 22$ $x=22$
If $y = - 3$ $y=-3$ , $x = 6$ $x=6$

Explanation:

$y^{2} = x + 3$ $y^2=x+3$ --- (1)
$x - 2 y = 12$ $x-2y=12$ --- (2)

From (2)
$x - 2 y = 12$ $x-2y=12$
$x = 12 + 2 y$ $x=12+2y$ --- (3)

Sub (3) into (1)
$y^{2} = 12 + 2 y + 3$ $y^2=12+2y+3$
$y^{2} = 15 + 2 y$ $y^2=15+2y$
$y^{2} - 2 y - 15 = 0$ $y^2-2y-15=0$
$(y - 5) (y + 3) = 0$ $(y-5)(y+3)=0$
$y = 5$ $y=5$ or $y = - 3$ $y=-3$

If $y = 5$ $y=5$ ,
$x = 12 + 2 y$ $x=12+2y$
$x = 12 + 10$ $x=12+10$
$x = 22$ $x=22$

If $y = - 3$ $y=-3$ ,
$x = 12 + 2 y$ $x=12+2y$
$x = 12 - 6$ $x=12-6$
$x = 6$ $x=6$

Answer 2 · 2018-06-10T12:57:57

$(6, - 3), (22, 5)$ $(6,-3),(22,5)$

Explanation:

$y^{2} = x + 3 \to (1)$ $y^2=x+3to(1)$

$x - 2 y = 12 \to (2)$ $x-2y=12to(2)$

$from equation (2) \to x = 12 + 2 y \to (3)$ $"from equation "(2)tox=12+2yto(3)$

$substitute x = 12 + 2 y in equation (1)$ $"substitute "x=12+2y" in equation "(1)$

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

$y^{2} - 2 y - 15 = 0 \leftarrow in standard form$ $y^2-2y-15=0larrcolor(blue)"in standard form"$

$(y - 5) (y + 3) = 0$ $(y-5)(y+3)=0$

$y + 3 = 0 \Rightarrow y = - 3$ $y+3=0rArry=-3$

$y - 5 = 0 \Rightarrow y = 5$ $y-5=0rArry=5$

$substitute these values into equation (3)$ $"substitute these values into equation "(3)$

$y = - 3 \to x = 12 - 6 = 6 \to (6, - 3)$ $y=-3tox=12-6=6to(6,-3)$

$y = 5 \to x = 12 + 10 = 22 \to (22, 5)$ $y=5tox=12+10=22to(22,5)$

Answer 3 · 2018-06-11T23:59:36

A graphical solution below:

Explanation:

Solving this graphically:

Here's one point of intersection:

graph{(y^2-x-3)(x-2y-12)=0}

And here's the other:

graph{(y^2-x-3)(x-2y-12)=0[10,30,0,10]}

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How do you solve y² = x + 3 and x - 2y = 12?

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