Solve {l}{{(2x+7)}=0}{y={(3x+5)}x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(2 x + 7)} = 0 }\\ { y = {(3 x + 5)} x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{77}{4} = 19\frac{1}{4} = 19.25$$

Solution Steps

Consider the first equation. Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$2x=-7$$

Divide both sides by 2.

$$x=-\frac{7}{2}$$

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

$$y=\left(3\left(-\frac{7}{2}\right)+5\right)\left(-\frac{7}{2}\right)$$

Multiply 3 and -\frac{7}{2} to get -\frac{21}{2}.

$$y=\left(-\frac{21}{2}+5\right)\left(-\frac{7}{2}\right)$$

Add -\frac{21}{2} and 5 to get -\frac{11}{2}.

$$y=-\frac{11}{2}\left(-\frac{7}{2}\right)$$

Multiply -\frac{11}{2} and -\frac{7}{2} to get \frac{77}{4}.

$$y=\frac{77}{4}$$

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

$$z=\frac{77}{4}$$

The system is now solved.

$$x=-\frac{7}{2} y=\frac{77}{4} z=\frac{77}{4}$$