Solve {l}{4x+1=0}{text{Solvefor}ytext{where}}{y=x^4+4x^3-6x}

$$\left. \begin{array} { l } { 4 x + 1 = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = x ^ {4} + 4 x ^ {3} - 6 x } \end{array} \right.$$

Solve for x, y

$$y = \frac{369}{256} = 1\frac{113}{256} = 1.44140625$$

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.

$$y=\left(-\frac{1}{4}\right)^{4}+4\left(-\frac{1}{4}\right)^{3}-6\left(-\frac{1}{4}\right)$$

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

$$y=\frac{1}{256}+4\left(-\frac{1}{4}\right)^{3}-6\left(-\frac{1}{4}\right)$$

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

$$y=\frac{1}{256}+4\left(-\frac{1}{64}\right)-6\left(-\frac{1}{4}\right)$$

Multiply 4 and -\frac{1}{64} to get -\frac{1}{16}.

$$y=\frac{1}{256}-\frac{1}{16}-6\left(-\frac{1}{4}\right)$$

Subtract \frac{1}{16} from \frac{1}{256} to get -\frac{15}{256}.

$$y=-\frac{15}{256}-6\left(-\frac{1}{4}\right)$$

Multiply -6 and -\frac{1}{4} to get \frac{3}{2}.

$$y=-\frac{15}{256}+\frac{3}{2}$$

Add -\frac{15}{256} and \frac{3}{2} to get \frac{369}{256}.

$$y=\frac{369}{256}$$

The system is now solved.

$$x=-\frac{1}{4} y=\frac{369}{256}$$