Solve {l}{f{(x)}=-6x+3}{g{(x)}=3x+21x^-3}{h=f{(-3)}}{i=h}{j=i}{k=j}{l=k}{m=l}{n=m}{text{Solvefor}otext{where}}{o=n}

$$\left. \begin{array} { l } { f {(x)} = -6 x + 3 }\\ { g {(x)} = 3 x + 21 x ^ {-3} }\\ { h = f {(-3)} }\\ { i = h }\\ { j = i }\\ { k = j }\\ { l = k }\\ { m = l }\\ { n = m }\\ { \text{Solve for } o \text{ where} } \\ { o = n } \end{array} \right.$$

Solve for f, x, g, h, j, k, l, m, n, o

$$o=i$$

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.

$$i=f\left(-3\right)$$

Divide both sides by -3.

$$\frac{i}{-3}=f$$

Divide i by -3 to get -\frac{1}{3}i.

$$-\frac{1}{3}i=f$$

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

$$f=-\frac{1}{3}i$$

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

$$-\frac{1}{3}ix=-6x+3$$

Add 6x to both sides.

$$-\frac{1}{3}ix+6x=3$$

Combine -\frac{1}{3}ix and 6x to get \left(6-\frac{1}{3}i\right)x.

$$\left(6-\frac{1}{3}i\right)x=3$$

Divide both sides by 6-\frac{1}{3}i.

$$x=\frac{3}{6-\frac{1}{3}i}$$

Multiply both numerator and denominator of \frac{3}{6-\frac{1}{3}i} by the complex conjugate of the denominator, 6+\frac{1}{3}i.

$$x=\frac{3\left(6+\frac{1}{3}i\right)}{\left(6-\frac{1}{3}i\right)\left(6+\frac{1}{3}i\right)}$$

Do the multiplications in \frac{3\left(6+\frac{1}{3}i\right)}{\left(6-\frac{1}{3}i\right)\left(6+\frac{1}{3}i\right)}.

$$x=\frac{18+i}{\frac{325}{9}}$$

Divide 18+i by \frac{325}{9} to get \frac{162}{325}+\frac{9}{325}i.

$$x=\frac{162}{325}+\frac{9}{325}i$$

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

$$g\left(\frac{162}{325}+\frac{9}{325}i\right)=3\left(\frac{162}{325}+\frac{9}{325}i\right)+21\left(\frac{162}{325}+\frac{9}{325}i\right)^{-3}$$

Multiply 3 and \frac{162}{325}+\frac{9}{325}i to get \frac{486}{325}+\frac{27}{325}i.

$$g\left(\frac{162}{325}+\frac{9}{325}i\right)=\frac{486}{325}+\frac{27}{325}i+21\left(\frac{162}{325}+\frac{9}{325}i\right)^{-3}$$

Calculate \frac{162}{325}+\frac{9}{325}i to the power of -3 and get \frac{214}{27}-\frac{971}{729}i.

$$g\left(\frac{162}{325}+\frac{9}{325}i\right)=\frac{486}{325}+\frac{27}{325}i+21\left(\frac{214}{27}-\frac{971}{729}i\right)$$

Multiply 21 and \frac{214}{27}-\frac{971}{729}i to get \frac{1498}{9}-\frac{6797}{243}i.

$$g\left(\frac{162}{325}+\frac{9}{325}i\right)=\frac{486}{325}+\frac{27}{325}i+\left(\frac{1498}{9}-\frac{6797}{243}i\right)$$

Add \frac{486}{325}+\frac{27}{325}i and \frac{1498}{9}-\frac{6797}{243}i to get \frac{491224}{2925}-\frac{2202464}{78975}i.

$$g\left(\frac{162}{325}+\frac{9}{325}i\right)=\frac{491224}{2925}-\frac{2202464}{78975}i$$

Divide both sides by \frac{162}{325}+\frac{9}{325}i.

$$g=\frac{\frac{491224}{2925}-\frac{2202464}{78975}i}{\frac{162}{325}+\frac{9}{325}i}$$

Multiply both numerator and denominator of \frac{\frac{491224}{2925}-\frac{2202464}{78975}i}{\frac{162}{325}+\frac{9}{325}i} by the complex conjugate of the denominator, \frac{162}{325}-\frac{9}{325}i.

$$g=\frac{\left(\frac{491224}{2925}-\frac{2202464}{78975}i\right)\left(\frac{162}{325}-\frac{9}{325}i\right)}{\left(\frac{162}{325}+\frac{9}{325}i\right)\left(\frac{162}{325}-\frac{9}{325}i\right)}$$

Do the multiplications in \frac{\left(\frac{491224}{2925}-\frac{2202464}{78975}i\right)\left(\frac{162}{325}-\frac{9}{325}i\right)}{\left(\frac{162}{325}+\frac{9}{325}i\right)\left(\frac{162}{325}-\frac{9}{325}i\right)}.

$$g=\frac{\frac{55984}{675}-\frac{18088}{975}i}{\frac{81}{325}}$$

Divide \frac{55984}{675}-\frac{18088}{975}i by \frac{81}{325} to get \frac{727792}{2187}-\frac{18088}{243}i.

$$g=\frac{727792}{2187}-\frac{18088}{243}i$$

The system is now solved.

$$f=-\frac{1}{3}i x=\frac{162}{325}+\frac{9}{325}i g=\frac{727792}{2187}-\frac{18088}{243}i h=i j=i k=i l=i m=i n=i o=i$$