Solve {l}{3z+3w=6}{3z+9{(3)}=6}{a=32}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 3 z + 3 w = 6 }\\ { 3 z + 9 {(3)} = 6 }\\ { a = 32 }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for z, w, a, b

$$b=32$$

Solution Steps

Consider the second equation. Multiply 9 and 3 to get 27.

$$3z+27=6$$

Subtract 27 from both sides.

$$3z=6-27$$

Subtract 27 from 6 to get -21.

$$3z=-21$$

Divide both sides by 3.

$$z=\frac{-21}{3}$$

Divide -21 by 3 to get -7.

$$z=-7$$

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

$$3\left(-7\right)+3w=6$$

Multiply 3 and -7 to get -21.

$$-21+3w=6$$

Add 21 to both sides.

$$3w=6+21$$

Add 6 and 21 to get 27.

$$3w=27$$

Divide both sides by 3.

$$w=\frac{27}{3}$$

Divide 27 by 3 to get 9.

$$w=9$$

The system is now solved.

$$z=-7 w=9 a=32 b=32$$