How do you solve $x y = 12$ $xy=12$ and $5 x - 3 y = - 8$ $5x-3y=-8$ ?

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Answer 1 · 2016-06-08T16:01:14

$y = \frac{12}{x}$ $y = 12/x$

$5 x - 3 (\frac{12}{x}) = - 8$ $5x - 3(12/x) = -8$

$5 x - \frac{36}{x} = - 8$ $5x - 36/x = -8$

$\frac{5 x^{2} - 36}{x} = - 8$ $(5x^2 - 36)/x = -8$

$5 x^{2} - 36 = - 8 x$ $5x^2 - 36 = -8x$

$5 x^{2} + 8 x - 36 = 0$ $5x^2 + 8x - 36 = 0$

$5 x^{2} - 10 x + 18 x - 36 = 0$ $5x^2 - 10x + 18x - 36 = 0$

$5 x (x - 2) + 18 (x - 2) = 0$ $5x(x - 2) + 18(x - 2) = 0$

$(5 x + 18) (x - 2) = 0$ $(5x + 18)(x - 2) = 0$

$x = - \frac{18}{5} and 2$ $x = -18/5 and 2$

$:. -18/5 xx y = 12 or 2 xx y = 12$

$y = 12/(-18/5) or y = 6$

$y = -10/3 or y = 6$

Our solution set is therefore ${-18/5, -10/3} and {2, 6}$ .

Hopefully this helps!

How do you solve xy=12xy=12xy=12 and 5x-3y=-85x−3y=−85x-3y=-8?