Solve {l}{1/3={(3-x)}}{y=4x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{1}{3} = {(3 - x)} }\\ { y = 4 x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{32}{3} = 10\frac{2}{3} \approx 10.666666667$$

Solution Steps

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$3-x=\frac{1}{3}$$

Subtract 3 from both sides.

$$-x=\frac{1}{3}-3$$

Subtract 3 from \frac{1}{3} to get -\frac{8}{3}.

$$-x=-\frac{8}{3}$$

Divide both sides by -1.

$$x=\frac{-\frac{8}{3}}{-1}$$

Express \frac{-\frac{8}{3}}{-1} as a single fraction.

$$x=\frac{-8}{3\left(-1\right)}$$

Multiply 3 and -1 to get -3.

$$x=\frac{-8}{-3}$$

Fraction \frac{-8}{-3} can be simplified to \frac{8}{3} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{8}{3}$$

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

$$y=4\times \frac{8}{3}$$

Multiply 4 and \frac{8}{3} to get \frac{32}{3}.

$$y=\frac{32}{3}$$

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

$$z=\frac{32}{3}$$

The system is now solved.

$$x=\frac{8}{3} y=\frac{32}{3} z=\frac{32}{3}$$