Solve {l}{x^3+7=y}{z^2+4^2=x}{z=sqrt{3}}

$$\left\{ \begin{array} { l } { x ^ { 3 } + 7 = y } \\ { z ^ { 2 } + 4 ^ { 2 } = x } \\ { z = \sqrt { 3 } } \end{array} \right.$$

Solve for x, y, z

$$x=19$$

$$y=6866$$

$$z=\sqrt{3}\approx 1.732050808$$

Solution Steps

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

$$\left(\sqrt{3}\right)^{2}+4^{2}=x$$

The square of \sqrt{3} is 3.

$$3+4^{2}=x$$

Calculate 4 to the power of 2 and get 16.

$$3+16=x$$

Add 3 and 16 to get 19.

$$19=x$$

Swap sides so that all variable terms are on the left hand side.

$$x=19$$

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

$$19^{3}+7=y$$

Calculate 19 to the power of 3 and get 6859.

$$6859+7=y$$

Add 6859 and 7 to get 6866.

$$6866=y$$

Swap sides so that all variable terms are on the left hand side.

$$y=6866$$

The system is now solved.

$$x=19 y=6866 z=\sqrt{3}$$