Solve {l}{x+y+z=210}{10x=30y}{10x=24z}

$$\left\{ \begin{array} { l } { x + y + z = 210 } \\ { 10 x = 30 y } \\ { 10 x = 24 z } \end{array} \right.$$

Solve for x, y, z

$$x=120$$

$$y=40$$

$$z=50$$

Short Steps Using Substitution

Solve x+y+z=210 for x.

$$x=-y-z+210$$

Substitute -y-z+210 for x in the second and third equation.

$$10\left(-y-z+210\right)=30y 10\left(-y-z+210\right)=24z$$

Solve these equations for y and z respectively.

$$y=\frac{105}{2}-\frac{1}{4}z z=\frac{1050}{17}-\frac{5}{17}y$$

Substitute \frac{105}{2}-\frac{1}{4}z for y in the equation z=\frac{1050}{17}-\frac{5}{17}y.

$$z=\frac{1050}{17}-\frac{5}{17}\left(\frac{105}{2}-\frac{1}{4}z\right)$$

Solve z=\frac{1050}{17}-\frac{5}{17}\left(\frac{105}{2}-\frac{1}{4}z\right) for z.

$$z=50$$

Substitute 50 for z in the equation y=\frac{105}{2}-\frac{1}{4}z.

$$y=\frac{105}{2}-\frac{1}{4}\times 50$$

Calculate y from y=\frac{105}{2}-\frac{1}{4}\times 50.

$$y=40$$

Substitute 40 for y and 50 for z in the equation x=-y-z+210.

$$x=-40-50+210$$

Calculate x from x=-40-50+210.

$$x=120$$

The system is now solved.

$$x=120 y=40 z=50$$