Solve {l}{6y+1=0}{text{Solvefor}ztext{where}}{z=y^4-6y^3+7y^2}

$$\left. \begin{array} { l } { 6 y + 1 = 0 }\\ { \text{Solve for } z \text{ where} } \\ { z = y ^ {4} - 6 y ^ {3} + 7 y ^ {2} } \end{array} \right.$$

Solve for y, z

$$z=\frac{289}{1296}\approx 0.222993827$$

Solution Steps

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

$$6y=-1$$

Divide both sides by 6.

$$y=-\frac{1}{6}$$

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

$$z=\left(-\frac{1}{6}\right)^{4}-6\left(-\frac{1}{6}\right)^{3}+7\left(-\frac{1}{6}\right)^{2}$$

Calculate -\frac{1}{6} to the power of 4 and get \frac{1}{1296}.

$$z=\frac{1}{1296}-6\left(-\frac{1}{6}\right)^{3}+7\left(-\frac{1}{6}\right)^{2}$$

Calculate -\frac{1}{6} to the power of 3 and get -\frac{1}{216}.

$$z=\frac{1}{1296}-6\left(-\frac{1}{216}\right)+7\left(-\frac{1}{6}\right)^{2}$$

Multiply -6 and -\frac{1}{216} to get \frac{1}{36}.

$$z=\frac{1}{1296}+\frac{1}{36}+7\left(-\frac{1}{6}\right)^{2}$$

Add \frac{1}{1296} and \frac{1}{36} to get \frac{37}{1296}.

$$z=\frac{37}{1296}+7\left(-\frac{1}{6}\right)^{2}$$

Calculate -\frac{1}{6} to the power of 2 and get \frac{1}{36}.

$$z=\frac{37}{1296}+7\times \frac{1}{36}$$

Multiply 7 and \frac{1}{36} to get \frac{7}{36}.

$$z=\frac{37}{1296}+\frac{7}{36}$$

Add \frac{37}{1296} and \frac{7}{36} to get \frac{289}{1296}.

$$z=\frac{289}{1296}$$

The system is now solved.

$$y=-\frac{1}{6} z=\frac{289}{1296}$$