Solve {l}{z=2}{a={(8/2)}+{(6/2)}*{(40/7)}}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { z = 2 }\\ { a = {(\frac{8}{2})} + {(\frac{6}{2})} \cdot {(\frac{40}{7})} }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for z, a, b, c, d

$$d = \frac{148}{7} = 21\frac{1}{7} \approx 21.142857143$$

Solution Steps

Consider the second equation. Divide 8 by 2 to get 4.

$$a=4+\frac{6}{2}\times \frac{40}{7}$$

Divide 6 by 2 to get 3.

$$a=4+3\times \frac{40}{7}$$

Multiply 3 and \frac{40}{7} to get \frac{120}{7}.

$$a=4+\frac{120}{7}$$

Add 4 and \frac{120}{7} to get \frac{148}{7}.

$$a=\frac{148}{7}$$

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

$$b=\frac{148}{7}$$

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

$$c=\frac{148}{7}$$

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

$$d=\frac{148}{7}$$

The system is now solved.

$$z=2 a=\frac{148}{7} b=\frac{148}{7} c=\frac{148}{7} d=\frac{148}{7}$$