Solve {l}{8x-2=0}{text{Solvefor}ytext{where}}{y=x^4+2x^3+5x^2}

$$\left. \begin{array} { l } { 8 x - 2 = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = x ^ {4} + 2 x ^ {3} + 5 x ^ {2} } \end{array} \right.$$

Solve for x, y

$$y=\frac{89}{256}=0.34765625$$

Solution Steps

Consider the first equation. Add 2 to both sides. Anything plus zero gives itself.

$$8x=2$$

Divide both sides by 8.

$$x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

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

$$y=\left(\frac{1}{4}\right)^{4}+2\times \left(\frac{1}{4}\right)^{3}+5\times \left(\frac{1}{4}\right)^{2}$$

Calculate \frac{1}{4} to the power of 4 and get \frac{1}{256}.

$$y=\frac{1}{256}+2\times \left(\frac{1}{4}\right)^{3}+5\times \left(\frac{1}{4}\right)^{2}$$

Calculate \frac{1}{4} to the power of 3 and get \frac{1}{64}.

$$y=\frac{1}{256}+2\times \frac{1}{64}+5\times \left(\frac{1}{4}\right)^{2}$$

Multiply 2 and \frac{1}{64} to get \frac{1}{32}.

$$y=\frac{1}{256}+\frac{1}{32}+5\times \left(\frac{1}{4}\right)^{2}$$

Add \frac{1}{256} and \frac{1}{32} to get \frac{9}{256}.

$$y=\frac{9}{256}+5\times \left(\frac{1}{4}\right)^{2}$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$y=\frac{9}{256}+5\times \frac{1}{16}$$

Multiply 5 and \frac{1}{16} to get \frac{5}{16}.

$$y=\frac{9}{256}+\frac{5}{16}$$

Add \frac{9}{256} and \frac{5}{16} to get \frac{89}{256}.

$$y=\frac{89}{256}$$

The system is now solved.

$$x=\frac{1}{4} y=\frac{89}{256}$$