Solve {l}{0.8x+{(0.2x)}*0.6=1}{y=0.6x-{(0.2x)}*0.8}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 0.8 x + {(0.2 x)} \cdot 0.6 = 1 }\\ { y = 0.6 x - {(0.2 x)} \cdot 0.8 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=\frac{11}{23}\approx 0.47826087$$

Solution Steps

Consider the first equation. Multiply 0.2 and 0.6 to get 0.12.

$$0.8x+0.12x=1$$

Combine 0.8x and 0.12x to get 0.92x.

$$0.92x=1$$

Divide both sides by 0.92.

$$x=\frac{1}{0.92}$$

Expand \frac{1}{0.92} by multiplying both numerator and the denominator by 100.

$$x=\frac{100}{92}$$

Reduce the fraction \frac{100}{92} to lowest terms by extracting and canceling out 4.

$$x=\frac{25}{23}$$

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

$$y=0.6\times \frac{25}{23}-0.2\times \frac{25}{23}\times 0.8$$

Multiply 0.6 and \frac{25}{23} to get \frac{15}{23}.

$$y=\frac{15}{23}-0.2\times \frac{25}{23}\times 0.8$$

Multiply 0.2 and \frac{25}{23} to get \frac{5}{23}.

$$y=\frac{15}{23}-\frac{5}{23}\times 0.8$$

Multiply \frac{5}{23} and 0.8 to get \frac{4}{23}.

$$y=\frac{15}{23}-\frac{4}{23}$$

Subtract \frac{4}{23} from \frac{15}{23} to get \frac{11}{23}.

$$y=\frac{11}{23}$$

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

$$z=\frac{11}{23}$$

The system is now solved.

$$x=\frac{25}{23} y=\frac{11}{23} z=\frac{11}{23}$$