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

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

Solve for x, y, z

$$x = -\frac{19}{5} = -3\frac{4}{5} = -3.8$$

$$y=-\frac{1}{5}=-0.2$$

$$z = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

Short Steps Using Substitution

Solve x+y+5z=3 for x.

$$x=-y-5z+3$$

Solve the second equation for y and the third equation for z.

$$y=-3z+4 z=-2y+1$$

Substitute -3z+4 for y in the equation z=-2y+1.

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

Solve z=-2\left(-3z+4\right)+1 for z.

$$z=\frac{7}{5}$$

Substitute \frac{7}{5} for z in the equation y=-3z+4.

$$y=-3\times \frac{7}{5}+4$$

Calculate y from y=-3\times \frac{7}{5}+4.

$$y=-\frac{1}{5}$$

Substitute -\frac{1}{5} for y and \frac{7}{5} for z in the equation x=-y-5z+3.

$$x=-\left(-\frac{1}{5}\right)-5\times \frac{7}{5}+3$$

Calculate x from x=-\left(-\frac{1}{5}\right)-5\times \frac{7}{5}+3.

$$x=-\frac{19}{5}$$

The system is now solved.

$$x=-\frac{19}{5} y=-\frac{1}{5} z=\frac{7}{5}$$