Solve {l}{x=-1}{y=3}{z=x+y^2}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { x = -1 }\\ { y = 3 }\\ { z = x + y ^ {2} }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c=8$$

Solution Steps

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

$$z=-1+3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$z=-1+9$$

Add -1 and 9 to get 8.

$$z=8$$

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

$$a=8$$

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

$$b=8$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=8$$

The system is now solved.

$$x=-1 y=3 z=8 a=8 b=8 c=8$$