Solve {l}{2x+y-2z=10}{3x+2y-2z=1}{5x+4y+3z=4}

$$\left. \begin{array} { l } { 2 x + y - 2 z = 10 } \\ { 3 x + 2 y - 2 z = 1 } \\ { 5 x + 4 y + 3 z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{137}{5} = 27\frac{2}{5} = 27.4$$

$$y = -\frac{182}{5} = -36\frac{2}{5} = -36.4$$

$$z = \frac{21}{5} = 4\frac{1}{5} = 4.2$$

Short Steps Using Substitution

Solve 2x+y-2z=10 for y.

$$y=10-2x+2z$$

Substitute 10-2x+2z for y in the second and third equation.

$$3x+2\left(10-2x+2z\right)-2z=1 5x+4\left(10-2x+2z\right)+3z=4$$

Solve these equations for x and z respectively.

$$x=19+2z z=-\frac{36}{11}+\frac{3}{11}x$$

Substitute 19+2z for x in the equation z=-\frac{36}{11}+\frac{3}{11}x.

$$z=-\frac{36}{11}+\frac{3}{11}\left(19+2z\right)$$

Solve z=-\frac{36}{11}+\frac{3}{11}\left(19+2z\right) for z.

$$z=\frac{21}{5}$$

Substitute \frac{21}{5} for z in the equation x=19+2z.

$$x=19+2\times \frac{21}{5}$$

Calculate x from x=19+2\times \frac{21}{5}.

$$x=\frac{137}{5}$$

Substitute \frac{137}{5} for x and \frac{21}{5} for z in the equation y=10-2x+2z.

$$y=10-2\times \frac{137}{5}+2\times \frac{21}{5}$$

Calculate y from y=10-2\times \frac{137}{5}+2\times \frac{21}{5}.

$$y=-\frac{182}{5}$$

The system is now solved.

$$x=\frac{137}{5} y=-\frac{182}{5} z=\frac{21}{5}$$