Solve {l}{y-2z+3=0}{-2x-3y+3=-2}{3x+y+3z=4}

$$\left\{ \begin{array} { l } { y - 2 z + 3 = 0 } \\ { - 2 x - 3 y + 3 = - 2 } \\ { 3 x + y + 3 z = 4 } \end{array} \right.$$

Solve for y, z, x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$y=4$$

$$z = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Short Steps Using Substitution

Solve y-2z+3=0 for y.

$$y=2z-3$$

Substitute 2z-3 for y in the second and third equation.

$$-2x-3\left(2z-3\right)+3=-2 3x+2z-3+3z=4$$

Solve these equations for z and x respectively.

$$z=\frac{7}{3}-\frac{1}{3}x x=-\frac{5}{3}z+\frac{7}{3}$$

Substitute \frac{7}{3}-\frac{1}{3}x for z in the equation x=-\frac{5}{3}z+\frac{7}{3}.

$$x=-\frac{5}{3}\left(\frac{7}{3}-\frac{1}{3}x\right)+\frac{7}{3}$$

Solve x=-\frac{5}{3}\left(\frac{7}{3}-\frac{1}{3}x\right)+\frac{7}{3} for x.

$$x=-\frac{7}{2}$$

Substitute -\frac{7}{2} for x in the equation z=\frac{7}{3}-\frac{1}{3}x.

$$z=\frac{7}{3}-\frac{1}{3}\left(-\frac{7}{2}\right)$$

Calculate z from z=\frac{7}{3}-\frac{1}{3}\left(-\frac{7}{2}\right).

$$z=\frac{7}{2}$$

Substitute \frac{7}{2} for z in the equation y=2z-3.

$$y=2\times \frac{7}{2}-3$$

Calculate y from y=2\times \frac{7}{2}-3.

$$y=4$$

The system is now solved.

$$y=4 z=\frac{7}{2} x=-\frac{7}{2}$$