Solve {l}{4x+1=0}{text{Solvefor}ytext{where}}{y=x^5+3x^2}

$$\left. \begin{array} { l } { 4 x + 1 = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = x ^ {5} + 3 x ^ {2} } \end{array} \right.$$

Solve for x, y

$$y=\frac{191}{1024}\approx 0.186523438$$

Solution Steps

Consider the first equation. Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$4x=-1$$

Divide both sides by 4.

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

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

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

Calculate -\frac{1}{4} to the power of 5 and get -\frac{1}{1024}.

$$y=-\frac{1}{1024}+3\left(-\frac{1}{4}\right)^{2}$$

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

$$y=-\frac{1}{1024}+3\times \frac{1}{16}$$

Multiply 3 and \frac{1}{16} to get \frac{3}{16}.

$$y=-\frac{1}{1024}+\frac{3}{16}$$

Add -\frac{1}{1024} and \frac{3}{16} to get \frac{191}{1024}.

$$y=\frac{191}{1024}$$

The system is now solved.

$$x=-\frac{1}{4} y=\frac{191}{1024}$$