Solve {l}{7x-8=0}{text{Solvefor}ytext{where}}{y=x^4+5x^3-2x^2}

$$\left. \begin{array} { l } { 7 x - 8 = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = x ^ {4} + 5 x ^ {3} - 2 x ^ {2} } \end{array} \right.$$

Solve for x, y

$$y = \frac{15744}{2401} = 6\frac{1338}{2401} \approx 6.557267805$$

Solution Steps

Consider the first equation. Add 8 to both sides. Anything plus zero gives itself.

$$7x=8$$

Divide both sides by 7.

$$x=\frac{8}{7}$$

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

$$y=\left(\frac{8}{7}\right)^{4}+5\times \left(\frac{8}{7}\right)^{3}-2\times \left(\frac{8}{7}\right)^{2}$$

Calculate \frac{8}{7} to the power of 4 and get \frac{4096}{2401}.

$$y=\frac{4096}{2401}+5\times \left(\frac{8}{7}\right)^{3}-2\times \left(\frac{8}{7}\right)^{2}$$

Calculate \frac{8}{7} to the power of 3 and get \frac{512}{343}.

$$y=\frac{4096}{2401}+5\times \frac{512}{343}-2\times \left(\frac{8}{7}\right)^{2}$$

Multiply 5 and \frac{512}{343} to get \frac{2560}{343}.

$$y=\frac{4096}{2401}+\frac{2560}{343}-2\times \left(\frac{8}{7}\right)^{2}$$

Add \frac{4096}{2401} and \frac{2560}{343} to get \frac{22016}{2401}.

$$y=\frac{22016}{2401}-2\times \left(\frac{8}{7}\right)^{2}$$

Calculate \frac{8}{7} to the power of 2 and get \frac{64}{49}.

$$y=\frac{22016}{2401}-2\times \frac{64}{49}$$

Multiply -2 and \frac{64}{49} to get -\frac{128}{49}.

$$y=\frac{22016}{2401}-\frac{128}{49}$$

Subtract \frac{128}{49} from \frac{22016}{2401} to get \frac{15744}{2401}.

$$y=\frac{15744}{2401}$$

The system is now solved.

$$x=\frac{8}{7} y=\frac{15744}{2401}$$