I want to show that $f(x)$ is bijective and calculate it's inverse.
Let $$f : \mathbf{R} \to \mathbf{R} $$ be defined by $f (x) = \frac{3x}{5} + 7$
I understand that a bijection must be injective and surjective but I don't understand how to show it for a function.
To show that $f(x)=\frac{3x}{5}+7$ is injective by the definition, note that $$\frac{3a}{5}+7=\frac{3b}{5}+7 \Rightarrow \frac{3a}{5}=\frac{3b}{5}\Rightarrow 3a=3b\Rightarrow a=b.$$ Therefore $f(x)$ is injective.
To find the inverse, trade $x$ and $y$ values in $y=\frac{3x}{5}+7$, and then solve for $y$: $$x=\frac{3y}{5}+7 \Rightarrow 5x=3y+35 \Rightarrow 3y=5x-35 \Rightarrow y=\frac{5x-35}{3}.$$
We can formally say that the inverse function of $f(x)$ is $$f^{-1}(x)=\frac{5x-35}{3}.$$
Regarding surjection: To show that the function is surjective, we want to show that for every real number $y$, there is a real number $x$ so that $f(x)=y$. A typical member of the domain $x$ is $x=\frac{5y-35}{3}$ (note we have not "traded" the $x,y$ values here from the original equation, but the process is the same). Since $y$ is a real number, certainly $\frac{5y-35}{3}$ is a real number. Now see that when we evaluate $f(x)$ at $x=\frac{5y-35}{3}$, we get
$$f\left(\frac{5y-35}{3}\right)=\frac{3\left(\frac{5y-35}{3}\right)}{5}+7=y.$$ This is also a real number. This means that for any real number $y$ we can always find an $x$ to "cover" that number.
Very specifically: You choose any real number $y$. I declare that $x=\frac{5y-35}{3}$ is the real value $x$ that covers that number so that $f(x)=y$, and because my value for $x$ is always defined for all real numbers $y$, we are safe. The $y$ values, or the "range", is "all used up", for lack of a more common phrase.