Solve {l}{10x=3}{10y=2}{z=10^x+y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 10 x = 3 }\\ { 10 y = 2 }\\ { z = 10 ^ {x + y} }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=\sqrt{10}\approx 3.16227766$$

Solution Steps

Consider the first equation. Divide both sides by 10.

$$x=\frac{3}{10}$$

Consider the second equation. Divide both sides by 10.

$$y=\frac{2}{10}$$

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

$$y=\frac{1}{5}$$

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

$$z=10^{\frac{3}{10}+\frac{1}{5}}$$

Add \frac{3}{10} and \frac{1}{5} to get \frac{1}{2}.

$$z=10^{\frac{1}{2}}$$

Reorder the terms.

$$z=\sqrt{10}$$

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

$$a=\sqrt{10}$$

The system is now solved.

$$x=\frac{3}{10} y=\frac{1}{5} z=\sqrt{10} a=\sqrt{10}$$