Solve {l}{3z+8y=53}{15z+20z-5z=12}

$$\left\{ \begin{array} { l } { 3 z + 8 y = 53 } \\ { 15 z + 20 z - 5 z = 12 } \end{array} \right.$$

Solve for z, y

$$y = \frac{259}{40} = 6\frac{19}{40} = 6.475$$

$$z=\frac{2}{5}=0.4$$

Solution Steps

Consider the second equation. Combine 15z and 20z to get 35z.

$$35z-5z=12$$

Combine 35z and -5z to get 30z.

$$30z=12$$

Divide both sides by 30.

$$z=\frac{12}{30}$$

Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.

$$z=\frac{2}{5}$$

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

$$3\times \frac{2}{5}+8y=53$$

Multiply 3 and \frac{2}{5} to get \frac{6}{5}.

$$\frac{6}{5}+8y=53$$

Subtract \frac{6}{5} from both sides.

$$8y=53-\frac{6}{5}$$

Subtract \frac{6}{5} from 53 to get \frac{259}{5}.

$$8y=\frac{259}{5}$$

Divide both sides by 8.

$$y=\frac{\frac{259}{5}}{8}$$

Express \frac{\frac{259}{5}}{8} as a single fraction.

$$y=\frac{259}{5\times 8}$$

Multiply 5 and 8 to get 40.

$$y=\frac{259}{40}$$

The system is now solved.

$$z=\frac{2}{5} y=\frac{259}{40}$$