How do you solve the system $x^{2} + 4 x - 4 y - 16 = 0$ $x^2+4x-4y-16=0$ and $- 2 x + y + 1 = 0$ $-2x+y+1=0$ ?

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Answer 1 · 2016-08-30T21:13:09

The solutions are $(6, 11)$ $(6, 11)$ and $(- 2, - 5)$ $(-2, -5)$

Solve for one of the variables. Let's solve for $y$ $y$ in the second equation.

$y = - 1 + 2 x$ $y = -1 + 2x$

Insert into the first equation:

$x^{2} + 4 x - 4 (- 1 + 2 x) - 16 = 0$ $x^2 + 4x - 4(-1 + 2x) - 16 = 0$

$x^{2} + 4 x + 4 - 8 x - 16 = 0$ $x^2 + 4x + 4 - 8x - 16 = 0$

$x^{2} - 4 x - 12 = 0$ $x^2 - 4x - 12 = 0$

$(x - 6) (x + 2) = 0$ $(x - 6)(x + 2) = 0$

$x = 6 and - 2$ $x = 6 and -2$

We can now substitute and solve for $y$ $y$ :

$y = 2 x - 1$ $y = 2x - 1$

$y = 2 (6) - 1 AND 2 (- 2) - 1$ $y = 2(6) - 1" AND "2(-2) - 1$

$y = 11 AND - 5$ $y = 11" AND "-5$

Hopefully this helps!

How do you solve the system x2+4x−4y−16=0x^2+4x-4y-16=0 and −2x+y+1=0-2x+y+1=0?