The solution to a system can usually be found by graphing, but graphing may not always be the most precise or the most efficient way to solve a system. 

Instead of solving by graphing, we can solve the system algebraically. Here is one way.

If we subtract $3p$ from each side of the first equation, $3p + q = 71$, we get an equivalent equation: $q= 71 - 3p$. Rewriting the original equation this way allows us to isolate the variable $q$. 

Because $q$ is equal to $71-3p$, we can substitute the expression $71-3p$ in the place of $q$ in the second equation. Doing this gives us an equation with only one variable, $p$, and makes it possible to find $p$.

$\begin{aligned} 2p - q &= 30 &\quad& \text {original equation} \\ 2p - (71 - 3p) &=30 &\quad& \text {substitute }71-3p \text{ for }q\\ 2p - 71 + 3p &=30 &\quad& \text {apply distributive property}\\ 5p - 71 &= 30 &\quad& \text {combine like terms}\\ 5p &= 101 &\quad& \text {add 71 to both sides}\\ p &= \dfrac{101}{5} &\quad& \text {divide both sides by 5} \\ p&=20.2 \end{aligned}$

Now that we know the value of $p$, we can find the value of $q$ by substituting 20.2 for $p$ in either of the original equations and solving the equation.

The solution to the system is the pair $p=20.2$ and $q=10.4$, or the point $(20.2, 10.4)$ on the graph. 

This method of solving a system of equations is called solving by substitution, because we substituted an expression for $q$ into the second equation.