How do you solve the following system: $8 x - 7 y = - 39, 7 x - 2 y = - 67$ $8x-7y=-39, 7x-2y= -67$ ?

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Answer 1 · 2017-07-05T12:00:14

By arranging equations. $x = - \frac{391}{33}$ $x=-391/33$ and $y = - \frac{263}{33}$ $y=-263/33$

Multiply the first equation by $- 2$ $-2$ and the second by $7$ $7$ :

$- 16 x + 14 y = 78$ $-16x + 14 y = 78$
$49 x - 14 y = - 469$ $49x - 14 y = -469$

Add these equations (provided above), yielding

$49 x - 16 x = - 391$ $49x-16x = -391$

$33 x = - 391$ $33x = -391$

$x = - \frac{391}{33}$ $x = -391/33$

When you put this value for $x$ $x$ (for instance in the first original equation), you will find $y$ $y$ .

$8 \times (- \frac{391}{33}) - 7 y = - 39$ $8 xx(-391/33) - 7y = -39$

$- 7 y = - 39 + \frac{3128}{33}$ $-7y = -39 + (3128)/33$

$- 7 y = \frac{- 1287 + 3128}{33}$ $-7y = (-1287 + 3128)/33$

$- 7 y = \frac{1841}{33}$ $-7y = 1841/33$

$y = - \frac{1841}{231}$ $y = -1841/231$

$y = - \frac{263}{33}$ $y= -263/33$

How do you solve the following system: 8x-7y=-39, 7x-2y= -67 8x−7y=−39,7x−2y=−67 8x-7y=-39, 7x-2y= -67 ?