$(H \circ M)(k) = H(M(k)) = H(3 k - 5)$

$H$ is going to do something to $3 k - 5$ and produce $5 k + 1$.

These are both linear functions, therefore $H(d)$ should be a linear function.

First $H$ should multiply $3 k - 5$ by $\answer {\frac {5}{3}}$ to produce a $5k$.

$$\frac {5}{3} (3 k - 5) = 5k - \frac {25}{3}$$

Now, we need $-\frac {25}{3}$ to become $1$. $H$ should add $\answer {\frac {28}{3}}$.

$$H(d) = \frac {5}{3} d + \frac {28}{3}$$

$\blacktriangleright$ Check:

$$(H \circ M)(k) = H(M(k)) = H(3 k - 5) = \frac {5}{3} (3 k - 5) + \frac {28}{3} = 5k - \frac {25}{3} + \frac {28}{3} = 5k + \frac {3}{3} = 5k + 1$$

$(H \circ M)(k) = H(M(k)) = 3 M(k) - 5$

These are both linear functions, therefore $M(k)$ should be a linear function.

$M$ is going to be multiplied by $3$, therefore $M$ should start with $M(p) = \answer {\frac {5}{3}}p$. That way, when it is multiplied by $3$, we will get $5p$.

Now, we need $3$ times something to be $1$. $M$ should add $\answer {\frac {1}{3}}$.

$$M(p) = \frac {5}{3} p + \frac {1}{3}$$

$\blacktriangleright$ Check:

$$(H \circ M)(k) = H(M(k)) = 3 \left ( \frac {5}{3} k + \frac {1}{3} \right ) - 5 = 5 k + 1$$

$(f \circ g)(t) = f(g(t)) = \frac {2t}{t-1}$

The composition has two occurrences of the variable. $g$ has one occurrence and when it is put into $f$ the resulting formula has two occurrences. It seems reasonable to guess that $f$ has two occurences of its variable - one in a numerator and one in a denominator. $3t + 1$ will replace each in the composition.

So, let’s start constructing.

$\blacktriangleright$ Guess #1)

Let’s start with two occurences of $x$, one in the numerator and one in the denominator.

$$f(x) = \frac {x}{x}$$

this makes the composition look like

$$f(g(t)) = \frac {3t + 1}{3t + 1}$$

We need the coefficients of $t$ to be $2$ in the numerator and $1$ in the denominator.

$\blacktriangleright$ Guess #2)

$$f(g(t)) = \frac {\answer {\frac {2}{3}}(3t + 1)}{\frac {1}{3}(3t + 1)}$$

That gives

$$f(g(t)) = \frac {2t + \frac {2}{3}}{t + \frac {1}{3}}$$

Now, we need constant terms of $0$ in the numerator and $-1$ in the denominator. Subtract $\answer {\frac {2}{3}}$ from the numerator. Subtract $\answer {\frac {4}{3}}$ from the denominator.

$\blacktriangleright$ Guess #3)

$$f(g(t)) = \frac {\frac {2}{3}(3t + 1) - \frac {2}{3}}{\frac {1}{3}(3t + 1) - \frac {4}{3}}$$

That tells us what $f$ needs to do.

$$f(x) = \frac {\frac {2}{3}x - \frac {2}{3}}{\frac {1}{3}x - \frac {4}{3}}$$

$(B \circ H)(t) = B(H(t)) = 3 H(t) + 1$

We want this to equal $\frac {2t}{t-1}$.

$$3 H(t) + 1 = \frac {2t}{t-1}$$

We can solve for $H(t)$.

$$3 H(t) + 1 = \frac {2t}{t-1}$$

$$3 H(t) = \frac {2t}{t-1} - 1 = \frac {\answer {t+1}}{t-1}$$

$$H(t) = \answer {\frac {t+1}{3(t-1)}}$$

Check:

$$(B \circ H)(t) = B(H(t)) = 3 H(t) + 1 = 3 \left ( \frac {t+1}{3(t-1)} \right ) + 1 = \frac {3(t+1) + 3(t-1)}{3(t-1)}$$

$$= \frac {(t+1) + (t-1)}{(t-1)} = \frac {2t}{(t-1)}$$