Solve {l}{x=7}{y=x{(x^2+3)}{(x^2-4)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = 7 }\\ { y = x {(x ^ {2} + 3)} {(x ^ {2} - 4)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=16380$$

Solution Steps

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

$$y=7\left(7^{2}+3\right)\left(7^{2}-4\right)$$

Calculate 7 to the power of 2 and get 49.

$$y=7\left(49+3\right)\left(7^{2}-4\right)$$

Add 49 and 3 to get 52.

$$y=7\times 52\left(7^{2}-4\right)$$

Multiply 7 and 52 to get 364.

$$y=364\left(7^{2}-4\right)$$

Calculate 7 to the power of 2 and get 49.

$$y=364\left(49-4\right)$$

Subtract 4 from 49 to get 45.

$$y=364\times 45$$

Multiply 364 and 45 to get 16380.

$$y=16380$$

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

$$z=16380$$

The system is now solved.

$$x=7 y=16380 z=16380$$