Solve {c}{8x+3y+10z=5}{-4x+9y-5z=1}{2x-12y-15z=-1}

$$\left. \begin{array} { c } { 8 x + 3 y + 10 z = 5 } \\ { - 4 x + 9 y - 5 z = 1 } \\ { 2 x - 12 y - 15 z = - 1 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{9}{14}\approx 0.642857143$$

$$y=\frac{1}{3}\approx 0.333333333$$

$$z=-\frac{4}{35}\approx -0.114285714$$

Short Steps Using Substitution

Solve 8x+3y+10z=5 for x.

$$x=-\frac{3}{8}y-\frac{5}{4}z+\frac{5}{8}$$

Substitute -\frac{3}{8}y-\frac{5}{4}z+\frac{5}{8} for x in the second and third equation.

$$-4\left(-\frac{3}{8}y-\frac{5}{4}z+\frac{5}{8}\right)+9y-5z=1 2\left(-\frac{3}{8}y-\frac{5}{4}z+\frac{5}{8}\right)-12y-15z=-1$$

Solve these equations for y and z respectively.

$$y=\frac{1}{3} z=\frac{9}{70}-\frac{51}{70}y$$

Substitute \frac{1}{3} for y in the equation z=\frac{9}{70}-\frac{51}{70}y.

$$z=\frac{9}{70}-\frac{51}{70}\times \frac{1}{3}$$

Calculate z from z=\frac{9}{70}-\frac{51}{70}\times \frac{1}{3}.

$$z=-\frac{4}{35}$$

Substitute \frac{1}{3} for y and -\frac{4}{35} for z in the equation x=-\frac{3}{8}y-\frac{5}{4}z+\frac{5}{8}.

$$x=-\frac{3}{8}\times \frac{1}{3}-\frac{5}{4}\left(-\frac{4}{35}\right)+\frac{5}{8}$$

Calculate x from x=-\frac{3}{8}\times \frac{1}{3}-\frac{5}{4}\left(-\frac{4}{35}\right)+\frac{5}{8}.

$$x=\frac{9}{14}$$

The system is now solved.

$$x=\frac{9}{14} y=\frac{1}{3} z=-\frac{4}{35}$$