Solve {l}{3x=30}{2y+x=20}{2z+y=13}

$$\left. \begin{array} { l } { 3 x = 30 } \\ { 2 y + x = 20 } \\ { 2 z + y = 13 } \end{array} \right.$$

Solve for x, y, z

$$x=10$$

$$y=5$$

$$z=4$$

Solution Steps

Consider the first equation. Divide both sides by 3.

$$x=\frac{30}{3}$$

Divide 30 by 3 to get 10.

$$x=10$$

Consider the second equation. Insert the known values of variables into the equation.

$$2y+10=20$$

Subtract 10 from both sides.

$$2y=20-10$$

Subtract 10 from 20 to get 10.

$$2y=10$$

Divide both sides by 2.

$$y=\frac{10}{2}$$

Divide 10 by 2 to get 5.

$$y=5$$

Consider the third equation. Insert the known values of variables into the equation.

$$2z+5=13$$

Subtract 5 from both sides.

$$2z=13-5$$

Subtract 5 from 13 to get 8.

$$2z=8$$

Divide both sides by 2.

$$z=\frac{8}{2}$$

Divide 8 by 2 to get 4.

$$z=4$$

The system is now solved.

$$x=10 y=5 z=4$$