Solve {l}{6y-52-14=0}{3y+2z-6=0}

$$\left. \begin{array} { l } { 6 y - 52 - 14 = 0 } \\ { 3 y + 2 z - 6 = 0 } \end{array} \right.$$

Solve for y, z

$$y=11$$

$$z = -\frac{27}{2} = -13\frac{1}{2} = -13.5$$

Solution Steps

Consider the first equation. Subtract 14 from -52 to get -66.

$$6y-66=0$$

Add 66 to both sides. Anything plus zero gives itself.

$$6y=66$$

Divide both sides by 6.

$$y=\frac{66}{6}$$

Divide 66 by 6 to get 11.

$$y=11$$

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

$$3\times 11+2z-6=0$$

Multiply 3 and 11 to get 33.

$$33+2z-6=0$$

Subtract 6 from 33 to get 27.

$$27+2z=0$$

Subtract 27 from both sides. Anything subtracted from zero gives its negation.

$$2z=-27$$

Divide both sides by 2.

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

The system is now solved.

$$y=11 z=-\frac{27}{2}$$