Solve {l}{6y=47}{text{Solvefor}ztext{where}}{z=frac{11-{(-4y-4)}}{3}}

$$\left. \begin{array} { l } { 6 y = 47 }\\ { \text{Solve for } z \text{ where} } \\ { z = \frac{11 - {(-4 y - 4)}}{3} } \end{array} \right.$$

Solve for y, z

$$z = \frac{139}{9} = 15\frac{4}{9} \approx 15.444444444$$

Solution Steps

Consider the first equation. Divide both sides by 6.

$$y=\frac{47}{6}$$

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

$$z=\frac{11-\left(-4\times \frac{47}{6}-4\right)}{3}$$

Multiply -4 and \frac{47}{6} to get -\frac{94}{3}.

$$z=\frac{11-\left(-\frac{94}{3}-4\right)}{3}$$

Subtract 4 from -\frac{94}{3} to get -\frac{106}{3}.

$$z=\frac{11-\left(-\frac{106}{3}\right)}{3}$$

Multiply -1 and -\frac{106}{3} to get \frac{106}{3}.

$$z=\frac{11+\frac{106}{3}}{3}$$

Add 11 and \frac{106}{3} to get \frac{139}{3}.

$$z=\frac{\frac{139}{3}}{3}$$

Express \frac{\frac{139}{3}}{3} as a single fraction.

$$z=\frac{139}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$z=\frac{139}{9}$$

The system is now solved.

$$y=\frac{47}{6} z=\frac{139}{9}$$