Solve {l}{3x+3y=25}{2x+7y-3z=19}{3x+2y-z=18}

$$\left\{ \begin{array} { l } { 3 x + 3 y = 25 } \\ { 2 x + 7 y - 3 z = 19 } \\ { 3 x + 2 y - z = 18 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{65}{12} = 5\frac{5}{12} \approx 5.416666667$$

$$y = \frac{35}{12} = 2\frac{11}{12} \approx 2.916666667$$

$$z = \frac{49}{12} = 4\frac{1}{12} \approx 4.083333333$$

Short Steps Using Substitution

Reorder the equations.

$$3x+2y-z=18 2x+7y-3z=19 3x+3y=25$$

Solve 3x+2y-z=18 for z.

$$z=3x+2y-18$$

Substitute 3x+2y-18 for z in the equation 2x+7y-3z=19.

$$2x+7y-3\left(3x+2y-18\right)=19$$

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

$$y=-35+7x x=-y+\frac{25}{3}$$

Substitute -35+7x for y in the equation x=-y+\frac{25}{3}.

$$x=-\left(-35+7x\right)+\frac{25}{3}$$

Solve x=-\left(-35+7x\right)+\frac{25}{3} for x.

$$x=\frac{65}{12}$$

Substitute \frac{65}{12} for x in the equation y=-35+7x.

$$y=-35+7\times \frac{65}{12}$$

Calculate y from y=-35+7\times \frac{65}{12}.

$$y=\frac{35}{12}$$

Substitute \frac{35}{12} for y and \frac{65}{12} for x in the equation z=3x+2y-18.

$$z=3\times \frac{65}{12}+2\times \frac{35}{12}-18$$

Calculate z from z=3\times \frac{65}{12}+2\times \frac{35}{12}-18.

$$z=\frac{49}{12}$$

The system is now solved.

$$x=\frac{65}{12} y=\frac{35}{12} z=\frac{49}{12}$$