Solve {r}{-2x+y+3z=9}{2y-z=8}{-2z=-4}

$$\left\{ \begin{array} { r } { - 2 x + y + 3 z = 9 } \\ { 2 y - z = 8 } \\ { - 2 z = - 4 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=5$$

$$z=2$$

Solution Steps

Consider the third equation. Divide both sides by -2.

$$z=\frac{-4}{-2}$$

Divide -4 by -2 to get 2.

$$z=2$$

Consider the second equation. Insert the known values of variables into the equation.

$$2y-2=8$$

Add 2 to both sides.

$$2y=8+2$$

Add 8 and 2 to get 10.

$$2y=10$$

Divide both sides by 2.

$$y=\frac{10}{2}$$

Divide 10 by 2 to get 5.

$$y=5$$

Consider the first equation. Insert the known values of variables into the equation.

$$-2x+5+3\times 2=9$$

Multiply 3 and 2 to get 6.

$$-2x+5+6=9$$

Add 5 and 6 to get 11.

$$-2x+11=9$$

Subtract 11 from both sides.

$$-2x=9-11$$

Subtract 11 from 9 to get -2.

$$-2x=-2$$

Divide both sides by -2.

$$x=\frac{-2}{-2}$$

Divide -2 by -2 to get 1.

$$x=1$$

The system is now solved.

$$x=1 y=5 z=2$$