We are given the system of equations:

1. $p + q + r = 4$
2. $-p + q + r = 2$

To solve for $p$, let's eliminate $q$ and $r$ by adding the two equations.

Step 1: Add the two equations

$(p + q + r) + (-p + q + r) = 4 + 2$

Simplifying both sides:

$p - p + q + q + r + r = 6$

$0 + 2q + 2r = 6$

$2(q + r) = 6$

Step 2: Solve for $q + r$

Now, divide both sides by 2:

$q + r = 3$

Step 3: Substitute $q + r = 3$ into the first equation

Substitute $q + r = 3$ into the first equation $p + q + r = 4$:

$p + 3 = 4$

Step 4: Solve for $p$

Now, subtract 3 from both sides:

$p = 1$

Thus, the value of $p$ is $\boxed{1}$.

Would you like further details on this solution or have any questions?

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Here are some related questions:

1. How would you solve for $q$ and $r$ if $p$ were known?
2. How can systems of linear equations be solved using substitution?
3. What happens if the two equations were inconsistent?
4. What other methods can be used to solve this system of equations?
5. Can we solve for $p$ if the two equations are multiplied instead of added?

Tip: When solving systems of linear equations, adding or subtracting equations can often eliminate one variable, making it easier to solve for the others.