How do you solve the system $16 x^{2} - y^{2} + 16 y - 128 = 0$ $16x^2-y^2+16y-128=0$ and $y^{2} - 48 x - 16 y - 32 = 0$ $y^2-48x-16y-32=0$ ?

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How do you solve the system $16 x^{2} - y^{2} + 16 y - 128 = 0$ $16x^2-y^2+16y-128=0$ and $y^{2} - 48 x - 16 y - 32 = 0$ $y^2-48x-16y-32=0$ ?

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Answer 1 · 2016-07-27T13:25:07

$(x, y) = (5, 8 \pm 4 \sqrt{21})$ $(x, y) = (5, 8+-4sqrt(21))$ or $(x, y) = (- 2, 8)$ $(x, y) = (-2, 8)$

Explanation:

If you add the two equations together, you get:

$16 x^{2} - 48 x - 160 = 0$ $16x^2-48x-160 = 0$

Divide through by $16$ $16$ to get:

$x^{2} - 3 x - 10 = 0$ $x^2-3x-10 = 0$

Find a pair of factors of $10$ $10$ which differ by $3$ $3$ . The pair $5, 2$ $5, 2$ works, so:

$0 = x^{2} - 3 x - 10 = (x - 5) (x + 2)$ $0 = x^2-3x-10 = (x-5)(x+2)$

Hence $x = 5$ $x=5$ or $x = - 2$ $x=-2$

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Substituting $x = 5$ $x=5$ in the second equation we get:

$0 = y^{2} - 16 y - 272$ $0 = y^2-16y-272$

$= {(y - 8)}^{2} - 336$ $= (y-8)^2-336$

$= {(y - 8)}^{2} - {(4 \sqrt{21})}^{2}$ $= (y-8)^2-(4sqrt(21))^2$

$= (y - 8 - 4 \sqrt{21}) (y - 8 + 4 \sqrt{21})$ $= (y-8-4sqrt(21))(y-8+4sqrt(21))$

So $y = 8 \pm 4 \sqrt{21}$ $y = 8 +-4sqrt(21)$

$color(white)()$
Substituting $x = - 2$ $x=-2$ in the second equation we get:

$0 = y^{2} - 16 + 64 = {(y - 8)}^{2}$ $0 = y^2-16+64 = (y-8)^2$

So $y = 8$ $y = 8$

Thus the solutions are:

$(x, y) = (5, 8 \pm 4 \sqrt{21})$ $(x, y) = (5, 8+-4sqrt(21))$ and $(x, y) = (- 2, 8)$ $(x, y) = (-2, 8)$

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How do you solve the system $16 x^{2} - y^{2} + 16 y - 128 = 0$ $16x^2-y^2+16y-128=0$ and $y^{2} - 48 x - 16 y - 32 = 0$ $y^2-48x-16y-32=0$ ?

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How do you solve the system 16x^2-y^2+16y-128=016x2−y2+16y−128=016x^2-y^2+16y-128=0 and y^2-48x-16y-32=0y2−48x−16y−32=0y^2-48x-16y-32=0?

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How do you solve the system $16 x^{2} - y^{2} + 16 y - 128 = 0$ $16x^2-y^2+16y-128=0$ and $y^{2} - 48 x - 16 y - 32 = 0$ $y^2-48x-16y-32=0$ ?