Solve {l}{x+y+5z=15}{y+3z=14}{2y=1}

$$\left\{ \begin{array}{l}{ x + y + 5 z = 15 }\\{ y + 3 z = 14 }\\{ 2 y = 1 }\end{array} \right.$$

Solve for x, y, z

$$x=-8$$

$$y=\frac{1}{2}=0.5$$

$$z = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Solution Steps

Consider the third equation. Divide both sides by 2.

$$y=\frac{1}{2}$$

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

$$\frac{1}{2}+3z=14$$

Subtract \frac{1}{2} from both sides.

$$3z=14-\frac{1}{2}$$

Subtract \frac{1}{2} from 14 to get \frac{27}{2}.

$$3z=\frac{27}{2}$$

Divide both sides by 3.

$$z=\frac{\frac{27}{2}}{3}$$

Express \frac{\frac{27}{2}}{3} as a single fraction.

$$z=\frac{27}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$z=\frac{27}{6}$$

Reduce the fraction \frac{27}{6} to lowest terms by extracting and canceling out 3.

$$z=\frac{9}{2}$$

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

$$x+\frac{1}{2}+5\times \frac{9}{2}=15$$

Multiply 5 and \frac{9}{2} to get \frac{45}{2}.

$$x+\frac{1}{2}+\frac{45}{2}=15$$

Add \frac{1}{2} and \frac{45}{2} to get 23.

$$x+23=15$$

Subtract 23 from both sides.

$$x=15-23$$

Subtract 23 from 15 to get -8.

$$x=-8$$

The system is now solved.

$$x=-8 y=\frac{1}{2} z=\frac{9}{2}$$