[b]p1.[/b] $ x,y,z$ are required to be non-negative whole numbers, find all solutions to the pair of equations $$x+y+z=40$$
$$2x + 4y + 17z = 301.$$


[b]p2.[/b] Let $P$ be a point lying between the sides of an acute angle whose vertex is $O$. Let $A,B$ be the intersections of a line passing through $P$ with the sides of the angle. Prove that the triangle $AOB$ has minimum area when $P$ bisects the line segment $AB$.


[b]p3.[/b] Find all values of $x$ for which $|3x-2|+|3x+1|=3$.


[b]p4.[/b] Prove that $x^2+y^2+z^2$ cannot be factored in the form $$(ax + by + cz) (dx + ey + fz),$$
$a, b, c, d, e, f$ real.


[b]p5.[/b] Let $f(x)$ be a continuous function for all real values of $x$ such that $f(a)\le f(b)$ whenever $a\le b$. Prove that, for every real number $r$, the equation $$x + f(x) = r$$ has exactly one solution.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the equation \( |3x-2| + |3x+1| = 3 \), we need to consider the different cases based on the values of \(3x-2\) and \(3x+1\). The critical points where the expressions inside the absolute values change sign are \( x = \frac{2}{3} \) and \( x = -\frac{1}{3} \).

1. **Case 1: \( x \geq \frac{2}{3} \)**
   - Here, both \(3x-2\) and \(3x+1\) are non-negative.
   - Therefore, \( |3x-2| = 3x-2 \) and \( |3x+1| = 3x+1 \).
   - The equation becomes:
     \[
     (3x-2) + (3x+1) = 3
     \]
     \[
     6x - 1 = 3
     \]
     \[
     6x = 4
     \]
     \[
     x = \frac{2}{3}
     \]
   - This solution is valid for \( x \geq \frac{2}{3} \).

2. **Case 2: \( -\frac{1}{3} \leq x < \frac{2}{3} \)**
   - Here, \(3x-2\) is negative and \(3x+1\) is positive.
   - Therefore, \( |3x-2| = -(3x-2) = -3x+2 \) and \( |3x+1| = 3x+1 \).
   - The equation becomes:
     \[
     (-3x+2) + (3x+1) = 3
     \]
     \[
     3 = 3
     \]
   - This is always true, so any \( x \) in the interval \( -\frac{1}{3} \leq x < \frac{2}{3} \) is a solution.

3. **Case 3: \( x < -\frac{1}{3} \)**
   - Here, both \(3x-2\) and \(3x+1\) are negative.
   - Therefore, \( |3x-2| = -(3x-2) = -3x+2 \) and \( |3x+1| = -(3x+1) = -3x-1 \).
   - The equation becomes:
     \[
     (-3x+2) + (-3x-1) = 3
     \]
     \[
     -6x + 1 = 3
     \]
     \[
     -6x = 2
     \]
     \[
     x = -\frac{1}{3}
     \]
   - This solution is valid for \( x < -\frac{1}{3} \).

Combining all the valid solutions from the cases, we get:
\[
x \in \left\{ -\frac{2}{3} \right\} \cup \left( -\frac{1}{3}, \frac{2}{3} \right]
\]

The final answer is \( \boxed{ x \in \left\{ -\frac{2}{3} \right\} \cup \left( -\frac{1}{3}, \frac{2}{3} \right] } \)