How do you solve the system of equations algebraically $x - y - z = 7, x + 2 y + 3 z = - 12, 3 x - 2 y + 7 z = 30$ $x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30$ ?

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How do you solve the system of equations algebraically $x - y - z = 7, x + 2 y + 3 z = - 12, 3 x - 2 y + 7 z = 30$ $x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30$ ?

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Answer 1 · 2017-03-29T11:58:49

Multiply the first equation by -3 first. Then go ahead, x=2/26; y=-226/26 and z=46/26

Explanation:

When you multiply the first equation by -3, you will get:
$- 3 x + 3 y + 3 z = - 21$ $-3x+3y+3z=-21$
Then combine this equation with the second. Yielding,
$y = 9 - 10 z$ $y=9-10z$ .

When you see y (in the first (original equation)), write this and for the second original equation do the same.

You will get $x + 9 z = 16$ $x+9z=16$ and $x - 17 z = - 30$ $x-17z=-30$ . You can multiply either the first equation by -1 or the second by -1. You will get
$z = \frac{46}{26}$ $z=46/26$

Since $y = 9 - 10 z$ $y=9-10z$ , you can find y now. $y = 9 - (\frac{460}{26})$ $y=9-(460/26)$ or $y = - \frac{226}{26}$ $y=-226/26$ .

Finally, you can easily solve using any original equation. $x = \frac{2}{26}$ $x=2/26$ .

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How do you solve the system of equations algebraically $x - y - z = 7, x + 2 y + 3 z = - 12, 3 x - 2 y + 7 z = 30$ $x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30$ ?

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How do you solve the system of equations algebraically x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30x−y−z=7,x+2y+3z=−12,3x−2y+7z=30x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30?

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How do you solve the system of equations algebraically $x - y - z = 7, x + 2 y + 3 z = - 12, 3 x - 2 y + 7 z = 30$ $x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30$ ?