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

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

Solve for x, y, z

$$z = -\frac{21}{4} = -5\frac{1}{4} = -5.25$$

Solution Steps

Consider the first equation. Combine 2x and 4x to get 6x.

$$6x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$6x=-3$$

Divide both sides by 6.

$$x=\frac{-3}{6}$$

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

$$x=-\frac{1}{2}$$

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

$$y=\left(-\frac{1}{2}+4\right)\left(-\frac{1}{2}-1\right)$$

Add -\frac{1}{2} and 4 to get \frac{7}{2}.

$$y=\frac{7}{2}\left(-\frac{1}{2}-1\right)$$

Subtract 1 from -\frac{1}{2} to get -\frac{3}{2}.

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

Multiply \frac{7}{2} and -\frac{3}{2} to get -\frac{21}{4}.

$$y=-\frac{21}{4}$$

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

$$z=-\frac{21}{4}$$

The system is now solved.

$$x=-\frac{1}{2} y=-\frac{21}{4} z=-\frac{21}{4}$$