Solve {c}{x+y+z=310}{45000x+40,000y+80,000z=16,000,000}{2x-y=0}

$$\left. \begin{array} { c } { x + y + z = 310 } \\ { 45000 x + 40,000 y + 80,000 z = 16,000,000 } \\ { 2 x - y = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{1760}{23} = 76\frac{12}{23} \approx 76.52173913$$

$$y = \frac{3520}{23} = 153\frac{1}{23} \approx 153.043478261$$

$$z = \frac{1850}{23} = 80\frac{10}{23} \approx 80.434782609$$

Short Steps Using Substitution

Solve x+y+z=310 for x.

$$x=-y-z+310$$

Substitute -y-z+310 for x in the second and third equation.

$$45000\left(-y-z+310\right)+40000y+80000z=16000000 2\left(-y-z+310\right)-y=0$$

Solve these equations for y and z respectively.

$$y=-410+7z z=310-\frac{3}{2}y$$

Substitute -410+7z for y in the equation z=310-\frac{3}{2}y.

$$z=310-\frac{3}{2}\left(-410+7z\right)$$

Solve z=310-\frac{3}{2}\left(-410+7z\right) for z.

$$z=\frac{1850}{23}$$

Substitute \frac{1850}{23} for z in the equation y=-410+7z.

$$y=-410+7\times \frac{1850}{23}$$

Calculate y from y=-410+7\times \frac{1850}{23}.

$$y=\frac{3520}{23}$$

Substitute \frac{3520}{23} for y and \frac{1850}{23} for z in the equation x=-y-z+310.

$$x=-\frac{3520}{23}-\frac{1850}{23}+310$$

Calculate x from x=-\frac{3520}{23}-\frac{1850}{23}+310.

$$x=\frac{1760}{23}$$

The system is now solved.

$$x=\frac{1760}{23} y=\frac{3520}{23} z=\frac{1850}{23}$$