Solve {l}{3x-y+2z=13}{4x+2y+5z=30}{5x-3y-z=3}

$$\left\{ \begin{array} { l } { 3 x - y + 2 z = 13 } \\ { 4 x + 2 y + 5 z = 30 } \\ { 5 x - 3 y - z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=4$$

Short Steps Using Substitution

Solve 3x-y+2z=13 for y.

$$y=3x+2z-13$$

Substitute 3x+2z-13 for y in the second and third equation.

$$4x+2\left(3x+2z-13\right)+5z=30 5x-3\left(3x+2z-13\right)-z=3$$

Solve these equations for x and z respectively.

$$x=-\frac{9}{10}z+\frac{28}{5} z=\frac{36}{7}-\frac{4}{7}x$$

Substitute -\frac{9}{10}z+\frac{28}{5} for x in the equation z=\frac{36}{7}-\frac{4}{7}x.

$$z=\frac{36}{7}-\frac{4}{7}\left(-\frac{9}{10}z+\frac{28}{5}\right)$$

Solve z=\frac{36}{7}-\frac{4}{7}\left(-\frac{9}{10}z+\frac{28}{5}\right) for z.

$$z=4$$

Substitute 4 for z in the equation x=-\frac{9}{10}z+\frac{28}{5}.

$$x=-\frac{9}{10}\times 4+\frac{28}{5}$$

Calculate x from x=-\frac{9}{10}\times 4+\frac{28}{5}.

$$x=2$$

Substitute 2 for x and 4 for z in the equation y=3x+2z-13.

$$y=3\times 2+2\times 4-13$$

Calculate y from y=3\times 2+2\times 4-13.

$$y=1$$

The system is now solved.

$$x=2 y=1 z=4$$