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The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function.

$F: mathbb{R} to mathbb{R}$ and $F^{−1}:mathbb{R} to mathbb{R}$ are defined by $F(x)=3x+2$ and $F^{−1}(y)=dfrac{y−2}{3}$. for all y ∈ R

My attempt:

Inverse Function

For each particular but arbitrarily chosen $y in mathbb{R}$, according to the definition of $f^{-1}$, $f^{-1}(y) = dfrac{y-2}{3}$ is a unique real number $x$ such that $f(x) = y$.
begin{align*}
F(x) & = y\
3x + 2 & = y\
x & = frac{y-2}{3}
end{align*}
Therefore, $f^{-1}(y) = frac{y-2}{3}$.

Compositions of Functions

The functions $g circ f$ and $f circ g$ are defined as follows:
$$(g circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2$$
for all $x in mathbb{Z}$.

Let's consider some definitions.

Definition. Let $A$, $B$, and $C$ be sets. Let $f: A to B$ and $g: B to C$ be functions. The composition of functions $g$ and $f$ is the function $g circ f: A to C$ defined by $(g circ f)(a) = g(f(a))$.

Definition. Let $f: A to B$ be a function. Then $g: B to A$ is said to be the inverse function of $f$ if
begin{align*}
(g circ f)(a) & = a & text{for each $a in A$}\
(f circ g)(b) & = b & text{for each $b in B$}
end{align*}
If $g$ is the inverse of $f$, we write $g = f^{-1}$, where $f^{-1}$ is read $f$ inverse.

Let $f: mathbb{R} to mathbb{R}$ be the function defined by $f(x) = 3x + 2$. Let $g: mathbb{R} to mathbb{R}$ be the function defined by $g(y) = frac{y - 2}{3}$. To show that $f$ and $g$ are inverses, we must show that $(g circ f)(x) = x$ for each $x in mathbb{R}$ and that $(f circ g)(y) = y$ for each $y in mathbb{R}$.

Let $x in mathbb{R}$. Then
begin{align*}
(g circ f)(x) & = g(f(x)) && text{by definition}\
& = g(3x + 2) && text{substitute $3x + 2$ for $f(x)$}\
& = frac{3x + 2 - 2}{3} && text{substitute $3x + 2$ for $x$ in the definition of $g$}\
& = frac{3x}{3} && text{simplify}\
& = x && text{simplify}
end{align*}
Hence, $(g circ f)(x) = x$ for each $x in mathbb{R}$.

It remains to show that $(f circ g)(y) = y$ for each $y in mathbb{R}$.

Let $y in mathbb{R}$. Then
begin{align*}
(f circ g)(y) & = f(g(y)) && text{by definition}\
& = fleft(frac{y - 2}{3}right) && text{substitute
$frac{y - 2}{3}$ for $g(y)$}\
& = 3left(frac{y - 2}{3}right) + 2 && text{substitute $frac{y - 2}{3}$ for $x$ in the definition of $f$}\
& = y - 2 + 2 && text{simplify}\
& = y && text{simplify}
end{align*}