Solve {l}{3+y-2=0}{6-3y+8z=33}

$$\left\{ \begin{array} { l } { 3 + y - 2 = 0 } \\ { 6 - 3 y + 8 z = 33 } \end{array} \right.$$

Solve for y, z

$$y=-1$$

$$z=3$$

Solution Steps

Consider the first equation. Subtract 2 from 3 to get 1.

$$1+y=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$y=-1$$

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

$$6-3\left(-1\right)+8z=33$$

Multiply -3 and -1 to get 3.

$$6+3+8z=33$$

Add 6 and 3 to get 9.

$$9+8z=33$$

Subtract 9 from both sides.

$$8z=33-9$$

Subtract 9 from 33 to get 24.

$$8z=24$$

Divide both sides by 8.

$$z=\frac{24}{8}$$

Divide 24 by 8 to get 3.

$$z=3$$

The system is now solved.

$$y=-1 z=3$$