Find the average rate of change of $g(t)=t^2+3t+1$ on the interval [0,a] I am working on a textbook question "Find the average rate of change of $g(t)=t^2+3t+1$ on the interval $[0,a]$". The solution provided, along with the steps in between is:
Avg. rate of change: 
= $\frac{g(a)-g(0)}{a-0}$
= $\frac{(a^2+3a+1)-(0^2+3(0)+1)}{a-0}$
= $\frac{a^2+3a+1-1}{a}$ # Here - how did they get rid of -0 in denominator?
= $\frac{a(a+3)}{a}$
= $a+3$
Where I'm stuck is between the second and third step. How does one go from denominator of $a-0$ to just $a$?

Just as $1-0=1$ or $5-0=5$ we get $a-0=a$ 
Note that $a$ represent an arbitrary real number so it follows real numbers axioms. 
