Solve {l}{x+x+x=60}{x+y=30}{x+z=3}

$$\left\{ \begin{array} { l } { x + x + x = 60 } \\ { x + y = 30 } \\ { x + z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x=20$$

$$y=10$$

$$z=-17$$

Solution Steps

Consider the first equation. Combine x and x to get 2x.

$$2x+x=60$$

Combine 2x and x to get 3x.

$$3x=60$$

Divide both sides by 3.

$$x=\frac{60}{3}$$

Divide 60 by 3 to get 20.

$$x=20$$

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

$$20+y=30$$

Subtract 20 from both sides.

$$y=30-20$$

Subtract 20 from 30 to get 10.

$$y=10$$

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

$$20+z=3$$

Subtract 20 from both sides.

$$z=3-20$$

Subtract 20 from 3 to get -17.

$$z=-17$$

The system is now solved.

$$x=20 y=10 z=-17$$