Solve {l}{x+1=3x}{y={(x^4+x^2)}}{text{Solvefor}ztext{where}}{z=y}

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

Solve for x, y, z

$$z=\frac{5}{16}=0.3125$$

Solution Steps

Consider the first equation. Subtract 3x from both sides.

$$x+1-3x=0$$

Combine x and -3x to get -2x.

$$-2x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-2x=-1$$

Divide both sides by -2.

$$x=\frac{-1}{-2}$$

Fraction \frac{-1}{-2} can be simplified to \frac{1}{2} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{1}{2}$$

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

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

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

$$y=\frac{1}{16}+\left(\frac{1}{2}\right)^{2}$$

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

$$y=\frac{1}{16}+\frac{1}{4}$$

Add \frac{1}{16} and \frac{1}{4} to get \frac{5}{16}.

$$y=\frac{5}{16}$$

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

$$z=\frac{5}{16}$$

The system is now solved.

$$x=\frac{1}{2} y=\frac{5}{16} z=\frac{5}{16}$$