Solve {l}{z={(-2)}}{a={(z^2+z)}+4}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { z = {(-2)} }\\ { a = {(z ^ {2} + z)} + 4 }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for z, a, b, c, d

$$d=6$$

Solution Steps

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

$$a=\left(-2\right)^{2}-2+4$$

Calculate -2 to the power of 2 and get 4.

$$a=4-2+4$$

Subtract 2 from 4 to get 2.

$$a=2+4$$

Add 2 and 4 to get 6.

$$a=6$$

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

$$b=6$$

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

$$c=6$$

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

$$d=6$$

The system is now solved.

$$z=-2 a=6 b=6 c=6 d=6$$