Solve {l}{x=10}{y=28}{z=(x+y)^2}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { x = 10 }\\ { y = 28 }\\ { z = {(x + y)} ^ {2} }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

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

$$c=1444$$

Solution Steps

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

$$z=\left(10+28\right)^{2}$$

Add 10 and 28 to get 38.

$$z=38^{2}$$

Calculate 38 to the power of 2 and get 1444.

$$z=1444$$

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

$$a=1444$$

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

$$b=1444$$

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

$$c=1444$$

The system is now solved.

$$x=10 y=28 z=1444 a=1444 b=1444 c=1444$$