Solve {l}{5x+4=5}{2x+3y=13}{z=0}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 5 x + 4 = 5 }\\ { 2 x + 3 y = 13 }\\ { z = 0 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=0$$

Solution Steps

Consider the first equation. Subtract 4 from both sides.

$$5x=5-4$$

Subtract 4 from 5 to get 1.

$$5x=1$$

Divide both sides by 5.

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

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

$$2\times \frac{1}{5}+3y=13$$

Multiply 2 and \frac{1}{5} to get \frac{2}{5}.

$$\frac{2}{5}+3y=13$$

Subtract \frac{2}{5} from both sides.

$$3y=13-\frac{2}{5}$$

Subtract \frac{2}{5} from 13 to get \frac{63}{5}.

$$3y=\frac{63}{5}$$

Divide both sides by 3.

$$y=\frac{\frac{63}{5}}{3}$$

Express \frac{\frac{63}{5}}{3} as a single fraction.

$$y=\frac{63}{5\times 3}$$

Multiply 5 and 3 to get 15.

$$y=\frac{63}{15}$$

Reduce the fraction \frac{63}{15} to lowest terms by extracting and canceling out 3.

$$y=\frac{21}{5}$$

The system is now solved.

$$x=\frac{1}{5} y=\frac{21}{5} z=0 a=0$$