Solve {c}{3x+y+2z=39}{x+2y+3z=26}{x=5+z}

$$\left. \begin{array} { c } { 3 x + y + 2 z = 39 } \\ { x + 2 y + 3 z = 26 } \\ { x = 5 + z } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{19}{2} = 9\frac{1}{2} = 9.5$$

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$z = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Short Steps Using Substitution

Reorder the equations.

$$x=5+z x+2y+3z=26 3x+y+2z=39$$

Substitute 5+z for x in the second and third equation.

$$5+z+2y+3z=26 3\left(5+z\right)+y+2z=39$$

Solve these equations for y and z respectively.

$$y=\frac{21}{2}-2z z=\frac{24}{5}-\frac{1}{5}y$$

Substitute \frac{21}{2}-2z for y in the equation z=\frac{24}{5}-\frac{1}{5}y.

$$z=\frac{24}{5}-\frac{1}{5}\left(\frac{21}{2}-2z\right)$$

Solve z=\frac{24}{5}-\frac{1}{5}\left(\frac{21}{2}-2z\right) for z.

$$z=\frac{9}{2}$$

Substitute \frac{9}{2} for z in the equation y=\frac{21}{2}-2z.

$$y=\frac{21}{2}-2\times \frac{9}{2}$$

Calculate y from y=\frac{21}{2}-2\times \frac{9}{2}.

$$y=\frac{3}{2}$$

Substitute \frac{9}{2} for z in the equation x=5+z.

$$x=5+\frac{9}{2}$$

Calculate x from x=5+\frac{9}{2}.

$$x=\frac{19}{2}$$

The system is now solved.

$$x=\frac{19}{2} y=\frac{3}{2} z=\frac{9}{2}$$