Solve {l}{{(2x+6)}=0}{y={(x+4)}{(1-4x)}+{(5+3x)}{(x+4)}+{(-x-4)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(2 x + 6)} = 0 }\\ { y = {(x + 4)} {(1 - 4 x)} + {(5 + 3 x)} {(x + 4)} + {(-x - 4)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=8$$

Solution Steps

Consider the first equation. Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x=-6$$

Divide both sides by 2.

$$x=\frac{-6}{2}$$

Divide -6 by 2 to get -3.

$$x=-3$$

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

$$y=\left(-3+4\right)\left(1-4\left(-3\right)\right)+\left(5+3\left(-3\right)\right)\left(-3+4\right)-\left(-3\right)-4$$

Add -3 and 4 to get 1.

$$y=1\left(1-4\left(-3\right)\right)+\left(5+3\left(-3\right)\right)\left(-3+4\right)-\left(-3\right)-4$$

Multiply -4 and -3 to get 12.

$$y=1\left(1+12\right)+\left(5+3\left(-3\right)\right)\left(-3+4\right)-\left(-3\right)-4$$

Add 1 and 12 to get 13.

$$y=1\times 13+\left(5+3\left(-3\right)\right)\left(-3+4\right)-\left(-3\right)-4$$

Multiply 1 and 13 to get 13.

$$y=13+\left(5+3\left(-3\right)\right)\left(-3+4\right)-\left(-3\right)-4$$

Multiply 3 and -3 to get -9.

$$y=13+\left(5-9\right)\left(-3+4\right)-\left(-3\right)-4$$

Subtract 9 from 5 to get -4.

$$y=13-4\left(-3+4\right)-\left(-3\right)-4$$

Add -3 and 4 to get 1.

$$y=13-4-\left(-3\right)-4$$

Subtract 4 from 13 to get 9.

$$y=9-\left(-3\right)-4$$

The opposite of -3 is 3.

$$y=9+3-4$$

Add 9 and 3 to get 12.

$$y=12-4$$

Subtract 4 from 12 to get 8.

$$y=8$$

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

$$z=8$$

The system is now solved.

$$x=-3 y=8 z=8$$