Solve {l}{f{(x)}={(x+3)}}{g=f{(3)}}{h=g}{i=h}{text{Solvefor}jtext{where}}{j=i}

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

Solve for f, x, g, h, j

$$j=i$$

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.

$$i=f\times 3$$

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.

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

Subtract x from both sides.

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

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

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

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

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

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

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

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

$$x=\frac{-3-i}{\frac{10}{9}}$$

Divide -3-i by \frac{10}{9} to get -\frac{27}{10}-\frac{9}{10}i.

$$x=-\frac{27}{10}-\frac{9}{10}i$$

The system is now solved.

$$f=\frac{1}{3}i x=-\frac{27}{10}-\frac{9}{10}i g=i h=i j=i$$