Solve {l}{24x=3.2+0.6x}{y=1}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 24 x = 3.2 + 0.6 x }\\ { y = 1 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=1$$

Solution Steps

Consider the first equation. Subtract 0.6x from both sides.

$$24x-0.6x=3.2$$

Combine 24x and -0.6x to get 23.4x.

$$23.4x=3.2$$

Divide both sides by 23.4.

$$x=\frac{3.2}{23.4}$$

Expand \frac{3.2}{23.4} by multiplying both numerator and the denominator by 10.

$$x=\frac{32}{234}$$

Reduce the fraction \frac{32}{234} to lowest terms by extracting and canceling out 2.

$$x=\frac{16}{117}$$

The system is now solved.

$$x=\frac{16}{117} y=1 z=1$$