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

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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

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

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

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

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

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

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=30?