Let's solve the given system of equations using the elimination method. The equations are:

1. $p + q - 40 = 0$
2. $7p + 3q = 180$

Step 1: Simplify the first equation

Rearrange the first equation to express it in standard form: $p + q = 40$

Step 2: Write both equations

Now, the system of equations is:

1. $p + q = 40$
2. $7p + 3q = 180$

Step 3: Eliminate one variable

We will eliminate $q$ by multiplying the first equation by $-3$ to make the coefficients of $q$ opposites: $-3(p + q) = -3(40)$ This simplifies to: $-3p - 3q = -120$

Now, our modified system of equations is:

1. $-3p - 3q = -120$
2. $7p + 3q = 180$

Step 4: Add the two equations

Add the two equations together to eliminate $q$: $(-3p - 3q) + (7p + 3q) = -120 + 180$ $4p = 60$ $p = 15$

Step 5: Substitute $p = 15$ into the first equation

Substitute $p = 15$ into the equation $p + q = 40$: $15 + q = 40$ $q = 25$

Conclusion

The values of $p$ and $q$ are: $p = 15, \quad q = 25$

The correct answer is: C) $p = 15, q = 25$

Would you like more details on the steps or have any questions?

Here are some related questions:

1. What is the elimination method and why is it useful in solving systems of equations?
2. How would you solve this system using the substitution method instead?
3. What are the conditions under which the elimination method is most effective?
4. Could this system have been solved graphically, and what would that look like?
5. What would happen if the equations represented parallel lines?

Tip: Always double-check your final values by substituting them back into the original equations to ensure they satisfy both conditions.