Solve {l}{1/2{(x-2/3)}=0}{y=2*1/{(frac{1*{(2)}+1}{2})}}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { \frac{1}{2} {(x - \frac{2}{3})} = 0 }\\ { y = 2 \cdot 1 / {(\frac{1 \cdot {(2)} + 1}{2})} }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Solution Steps

Consider the first equation. Multiply both sides by 2, the reciprocal of \frac{1}{2}. Anything times zero gives zero.

$$x-\frac{2}{3}=0$$

Add \frac{2}{3} to both sides. Anything plus zero gives itself.

$$x=\frac{2}{3}$$

Consider the second equation. Multiply 2 and 1 to get 2.

$$y=\frac{2}{\frac{1\times 2+1}{2}}$$

Multiply 1 and 2 to get 2.

$$y=\frac{2}{\frac{2+1}{2}}$$

Add 2 and 1 to get 3.

$$y=\frac{2}{\frac{3}{2}}$$

Divide 2 by \frac{3}{2} by multiplying 2 by the reciprocal of \frac{3}{2}.

$$y=2\times \frac{2}{3}$$

Multiply 2 and \frac{2}{3} to get \frac{4}{3}.

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

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

$$z=\frac{4}{3}$$

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

$$a=\frac{4}{3}$$

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

$$b=\frac{4}{3}$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=\frac{4}{3}$$

The system is now solved.

$$x=\frac{2}{3} y=\frac{4}{3} z=\frac{4}{3} a=\frac{4}{3} b=\frac{4}{3} c=\frac{4}{3}$$