How do you solve the system of equations $2 x - y + 3 z = - 1$ $2x-y+3z=-1$ , $x + 2 y - 4 z = - 1$ $x+2y-4z=-1$ and $y - 2 z = 0$ $y-2z=0$ by elimination?

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Answer 1 · 2017-09-07T17:23:21

From third equation $y = 2 z$ $y=2z$

After plugging y into second one,

$x + 2 \cdot 2 z - 4 z = - 1$ $x+2*2z-4z=-1$ or $x = - 1$ $x=-1$ .

After plugging x and y into first one,

$2 \cdot (- 1) - 2 z + 3 z = - 1$ $2*(-1)-2z+3z=-1$

$- 2 + z = - 1$ $-2+z=-1$

$z = 1$ $z=1$ . Consequently, $y = 2 \cdot 1 = 2$ $y=2*1=2$

1) I found y in terms of z from third equation.

2) I plugged y into second one for finding x.

3) I plugged x and y into first one for finding z.

4) I found y.

How do you solve the system of equations 2x−y+3z=−12x-y+3z=-1, x+2y−4z=−1x+2y-4z=-1 and y−2z=0y-2z=0 by elimination?