Solve {c}{2x+4y-2=9}{3x+y+2z=7}{x+3y-32=4}

$$\left. \begin{array} { c } { 2 x + 4 y - 2 = 9 } \\ { 3 x + y + 2 z = 7 } \\ { x + 3 y - 32 = 4 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{111}{2} = -55\frac{1}{2} = -55.5$$

$$y = \frac{61}{2} = 30\frac{1}{2} = 30.5$$

$$z = \frac{143}{2} = 71\frac{1}{2} = 71.5$$

Short Steps Using Substitution

Reorder the equations.

$$3x+y+2z=7 2x+4y-2=9 x+3y-32=4$$

Solve 3x+y+2z=7 for y.

$$y=-3x-2z+7$$

Substitute -3x-2z+7 for y in the second and third equation.

$$2x+4\left(-3x-2z+7\right)-2=9 x+3\left(-3x-2z+7\right)-32=4$$

Solve these equations for x and z respectively.

$$x=\frac{17}{10}-\frac{4}{5}z z=-\frac{5}{2}-\frac{4}{3}x$$

Substitute \frac{17}{10}-\frac{4}{5}z for x in the equation z=-\frac{5}{2}-\frac{4}{3}x.

$$z=-\frac{5}{2}-\frac{4}{3}\left(\frac{17}{10}-\frac{4}{5}z\right)$$

Solve z=-\frac{5}{2}-\frac{4}{3}\left(\frac{17}{10}-\frac{4}{5}z\right) for z.

$$z=\frac{143}{2}$$

Substitute \frac{143}{2} for z in the equation x=\frac{17}{10}-\frac{4}{5}z.

$$x=\frac{17}{10}-\frac{4}{5}\times \frac{143}{2}$$

Calculate x from x=\frac{17}{10}-\frac{4}{5}\times \frac{143}{2}.

$$x=-\frac{111}{2}$$

Substitute -\frac{111}{2} for x and \frac{143}{2} for z in the equation y=-3x-2z+7.

$$y=-3\left(-\frac{111}{2}\right)-2\times \frac{143}{2}+7$$

Calculate y from y=-3\left(-\frac{111}{2}\right)-2\times \frac{143}{2}+7.

$$y=\frac{61}{2}$$

The system is now solved.

$$x=-\frac{111}{2} y=\frac{61}{2} z=\frac{143}{2}$$