How do you solve the following system?: $7 x - 3 y = 11, - 9 x - 24 y = - 7$ $7x -3y =11, -9x -24y = -7$

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Answer 1 · 2018-06-24T15:22:52

$x = \frac{19}{13}$ $x=19/13$ and $y = - \frac{10}{39}$ $y=-10/39$

$8 \cdot (7 x - 3 y) - (- 9 x - 24 y) = 8 \cdot 11 - (- 7)$ $8*(7x-3y)-(-9x-24y)=8*11-(-7)$

$56 x - 24 y + 9 x + 24 y = 88 + 7$ $56x-24y+9x+24y=88+7$

$65 x = 95$ $65x=95$ , so $x = \frac{95}{65} = \frac{19}{13}$ $x=95/65=19/13$

Thus,

$7 \cdot \frac{19}{13} - 3 y = 11$ $7*19/13-3y=11$

$\frac{133}{13} - 3 y = 11$ $133/13-3y=11$

$- 3 y = \frac{10}{13}$ $-3y=10/13$ or $y = - \frac{10}{39}$ $y=-10/39$

Answer 2 · 2018-06-24T15:24:58

$x = \frac{19}{13}, y = - \frac{10}{39}$ $x=19/13,y=-10/39$

Multiplying the first equation by $- 8$ $-8$ and adding to the second we get

$- 65 x = - 95$ $-65x=-95$
so
$x = \frac{95}{65} = \frac{19}{13}$ $x=95/65=19/13$
plugging this in the first equation we get

$7 \cdot \frac{19}{13} - 3 y = 11$ $7*19/13-3y=11$
$\frac{133}{13} - 11 = 3 y$ $133/13-11=3y$

$\frac{133 - 143}{13} = 3 y$ $(133-143)/13=3y$

$- \frac{10}{13} = 3 y$ $-10/13=3y$

so

$y = - \frac{10}{39}$ $y=-10/39$

How do you solve the following system?: 7x−3y=11,−9x−24y=−77x -3y =11, -9x -24y = -7