Solve {l}{2x-7+4z=23}{3x+2y+3z=12}{x+6y-5z=-36}

$$\left. \begin{array} { l } { 2 x - 7 + 4 z = 23 } \\ { 3 x + 2 y + 3 z = 12 } \\ { x + 6 y - 5 z = - 36 } \end{array} \right.$$

Solve for x, z, y

$$x=-33$$

$$y = \frac{39}{2} = 19\frac{1}{2} = 19.5$$

$$z=24$$

Short Steps Using Substitution

Reorder the equations.

$$x+6y-5z=-36 3x+2y+3z=12 2x-7+4z=23$$

Solve x+6y-5z=-36 for x.

$$x=-6y+5z-36$$

Substitute -6y+5z-36 for x in the second and third equation.

$$3\left(-6y+5z-36\right)+2y+3z=12 2\left(-6y+5z-36\right)-7+4z=23$$

Solve these equations for z and y respectively.

$$z=\frac{20}{3}+\frac{8}{9}y y=-\frac{17}{2}+\frac{7}{6}z$$

Substitute \frac{20}{3}+\frac{8}{9}y for z in the equation y=-\frac{17}{2}+\frac{7}{6}z.

$$y=-\frac{17}{2}+\frac{7}{6}\left(\frac{20}{3}+\frac{8}{9}y\right)$$

Solve y=-\frac{17}{2}+\frac{7}{6}\left(\frac{20}{3}+\frac{8}{9}y\right) for y.

$$y=\frac{39}{2}$$

Substitute \frac{39}{2} for y in the equation z=\frac{20}{3}+\frac{8}{9}y.

$$z=\frac{20}{3}+\frac{8}{9}\times \frac{39}{2}$$

Calculate z from z=\frac{20}{3}+\frac{8}{9}\times \frac{39}{2}.

$$z=24$$

Substitute 24 for z and \frac{39}{2} for y in the equation x=-6y+5z-36.

$$x=-6\times \frac{39}{2}+5\times 24-36$$

Calculate x from x=-6\times \frac{39}{2}+5\times 24-36.

$$x=-33$$

The system is now solved.

$$x=-33 z=24 y=\frac{39}{2}$$