Solve {l}{f{(x)}=20{(2x^3+3x^2-2x)}}{g=8x}{h=g}{i=h}{j=i}{text{Solvefor}ktext{where}}{k=j}

$$\left. \begin{array} { l } { f {(x)} = 20 {(2 x ^ {3} + 3 x ^ {2} - 2 x)} }\\ { g = 8 x }\\ { h = g }\\ { i = h }\\ { j = i }\\ { \text{Solve for } k \text{ where} } \\ { k = j } \end{array} \right.$$

Solve for f, x, g, h, j, k

$$k=i$$

Solution Steps

Consider the fourth equation. Swap sides so that all variable terms are on the left hand side.

$$h=i$$

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

$$i=g$$

Swap sides so that all variable terms are on the left hand side.

$$g=i$$

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

$$i=8x$$

Divide both sides by 8.

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

Divide i by 8 to get \frac{1}{8}i.

$$\frac{1}{8}i=x$$

Swap sides so that all variable terms are on the left hand side.

$$x=\frac{1}{8}i$$

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

$$f\times \left(\frac{1}{8}i\right)=20\left(2\times \left(\frac{1}{8}i\right)^{3}+3\times \left(\frac{1}{8}i\right)^{2}-2\times \left(\frac{1}{8}i\right)\right)$$

Calculate \frac{1}{8}i to the power of 3 and get -\frac{1}{512}i.

$$f\times \left(\frac{1}{8}i\right)=20\left(2\times \left(-\frac{1}{512}i\right)+3\times \left(\frac{1}{8}i\right)^{2}-2\times \left(\frac{1}{8}i\right)\right)$$

Multiply 2 and -\frac{1}{512}i to get -\frac{1}{256}i.

$$f\times \left(\frac{1}{8}i\right)=20\left(-\frac{1}{256}i+3\times \left(\frac{1}{8}i\right)^{2}-2\times \left(\frac{1}{8}i\right)\right)$$

Calculate \frac{1}{8}i to the power of 2 and get -\frac{1}{64}.

$$f\times \left(\frac{1}{8}i\right)=20\left(-\frac{1}{256}i+3\left(-\frac{1}{64}\right)-2\times \left(\frac{1}{8}i\right)\right)$$

Multiply 3 and -\frac{1}{64} to get -\frac{3}{64}.

$$f\times \left(\frac{1}{8}i\right)=20\left(-\frac{1}{256}i-\frac{3}{64}-2\times \left(\frac{1}{8}i\right)\right)$$

Multiply -2 and \frac{1}{8}i to get -\frac{1}{4}i.

$$f\times \left(\frac{1}{8}i\right)=20\left(-\frac{1}{256}i-\frac{3}{64}-\frac{1}{4}i\right)$$

Do the additions in -\frac{1}{256}i-\frac{3}{64}-\frac{1}{4}i.

$$f\times \left(\frac{1}{8}i\right)=20\left(-\frac{3}{64}-\frac{65}{256}i\right)$$

Multiply 20 and -\frac{3}{64}-\frac{65}{256}i to get -\frac{15}{16}-\frac{325}{64}i.

$$f\times \left(\frac{1}{8}i\right)=-\frac{15}{16}-\frac{325}{64}i$$

Divide both sides by \frac{1}{8}i.

$$f=\frac{-\frac{15}{16}-\frac{325}{64}i}{\frac{1}{8}i}$$

Multiply both numerator and denominator of \frac{-\frac{15}{16}-\frac{325}{64}i}{\frac{1}{8}i} by imaginary unit i.

$$f=\frac{\frac{325}{64}-\frac{15}{16}i}{-\frac{1}{8}}$$

Divide \frac{325}{64}-\frac{15}{16}i by -\frac{1}{8} to get -\frac{325}{8}+\frac{15}{2}i.

$$f=-\frac{325}{8}+\frac{15}{2}i$$

The system is now solved.

$$f=-\frac{325}{8}+\frac{15}{2}i x=\frac{1}{8}i g=i h=i j=i k=i$$