Solve {l}{2/3x+1/5=2/10}{y=0-{(-8)}}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { \frac{2}{3} x + \frac{1}{5} = \frac{2}{10} }\\ { y = 0 - {(-8)} }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

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

$$c=8$$

Solution Steps

Consider the first equation. Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}x+\frac{1}{5}=\frac{1}{5}$$

Subtract \frac{1}{5} from both sides.

$$\frac{2}{3}x=\frac{1}{5}-\frac{1}{5}$$

Subtract \frac{1}{5} from \frac{1}{5} to get 0.

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

Multiply both sides by \frac{3}{2}, the reciprocal of \frac{2}{3}. Anything times zero gives zero.

$$x=0$$

Consider the second equation. The opposite of -8 is 8.

$$y=0+8$$

Add 0 and 8 to get 8.

$$y=8$$

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

$$z=8$$

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

$$a=8$$

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

$$b=8$$

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

$$c=8$$

The system is now solved.

$$x=0 y=8 z=8 a=8 b=8 c=8$$