Solve {l}{f{(t)}=3t+3/5}{g=f{(5)}}{h=g}{i=h}{j=i}{text{Solvefor}ktext{where}}{k=j}

$$\left. \begin{array} { l } { f {(t)} = \frac{3 t + 3}{5} }\\ { g = f {(5)} }\\ { h = g }\\ { i = h }\\ { j = i }\\ { \text{Solve for } k \text{ where} } \\ { k = j } \end{array} \right.$$

Solve for f, t, 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.

$$i=f\times 5$$

Divide both sides by 5.

$$\frac{i}{5}=f$$

Divide i by 5 to get \frac{1}{5}i.

$$\frac{1}{5}i=f$$

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

$$f=\frac{1}{5}i$$

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

$$\frac{1}{5}it=\frac{3t+3}{5}$$

Multiply both sides of the equation by 5.

$$it=3t+3$$

Subtract 3t from both sides.

$$it-3t=3$$

Combine it and -3t to get \left(-3+i\right)t.

$$\left(-3+i\right)t=3$$

Divide both sides by -3+i.

$$t=\frac{3}{-3+i}$$

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

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

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

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

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

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

The system is now solved.

$$f=\frac{1}{5}i t=-\frac{9}{10}-\frac{3}{10}i g=i h=i j=i k=i$$