Solve {l}{f{(x)}=x^3+x+4}{x=2}{g=0*3}{text{Solvefor}htext{where}}{h=g}

$$\left. \begin{array} { l } { f {(x)} = x ^ {3} + x + 4 }\\ { x = 2 }\\ { g = 0 \cdot 3 }\\ { \text{Solve for } h \text{ where} } \\ { h = g } \end{array} \right.$$

Solve for f, x, g, h

$$h=0$$

Solution Steps

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

$$f\times 2=2^{3}+2+4$$

Calculate 2 to the power of 3 and get 8.

$$f\times 2=8+2+4$$

Add 8 and 2 to get 10.

$$f\times 2=10+4$$

Add 10 and 4 to get 14.

$$f\times 2=14$$

Divide both sides by 2.

$$f=\frac{14}{2}$$

Divide 14 by 2 to get 7.

$$f=7$$

Consider the third equation. Multiply 0 and 3 to get 0.

$$g=0$$

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

$$h=0$$

The system is now solved.

$$f=7 x=2 g=0 h=0$$