Show that the function given by $f(x) = 7x - 3$ is strictly increasing on $R$.

(A) Let $x_{1}$ and $x_{2}$ be any two real numbers in $R$ such that $x_{1} < x_{2}$.
Multiplying both sides by $7$,we get $7x_{1} < 7x_{2}$.
Subtracting $3$ from both sides,we get $7x_{1} - 3 < 7x_{2} - 3$.
This implies $f(x_{1}) < f(x_{2})$.
Since $x_{1} < x_{2}$ implies $f(x_{1}) < f(x_{2})$ for all $x_{1}, x_{2} \in R$,the function $f(x) = 7x - 3$ is strictly increasing on $R$.
Alternatively,using the derivative test: $f'(x) = \frac{d}{dx}(7x - 3) = 7$.
Since $f'(x) = 7 > 0$ for all $x \in R$,the function is strictly increasing on $R$.