How do you solve the following system?: $- 12 x - 15 y = 3, 13 x - 16 y = - 4$ $-12x -15y =3, 13x -16y = -4$

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Answer 1 · 2017-07-24T11:26:49

After some arrangement, $x = - 12$ $x=-12$ and $y = \frac{1}{43}$ $y=1/43$

Enlarge the first equation with 13 and the second equation with 12:

$- 156 x - 195 y = 39$ $-156x - 195y = 39$
$156 x - 192 y = - 48$ $156x - 192y = -48$

Combine these two equations:

$- 387 y = - 9$ $-387y = -9$

$y = \frac{9}{387}$ $y= 9/387$

$y = \frac{1}{43}$ $y=1/43$

Now you can calculate x (for example using the second original equation):

$13 x - \frac{16}{43} = - 4$ $13x - 16/43 = -4$

$13 x = - 4 + \frac{16}{43}$ $13x = -4 + 16/43$

$13 x = - 156$ $13x = -156$

$x = - \frac{156}{13}$ $x= -156/13$

$x = - 12$ $x= -12$

How do you solve the following system?: -12x -15y =3, 13x -16y = -4−12x−15y=3,13x−16y=−4-12x -15y =3, 13x -16y = -4