The system x > m + 2 and -2x - 1 >= 4m + 1 has no solution, and the equation x + m - 2 = 2 - x has a non-negative integer solution. Find the sum of all integers m that satisfy both conditions.

Step-by-step solution

1. 1

   Rewrite the second inequality with x isolated. From $-2x-1\ge 4m+1$ we get $-2x\ge 4m+2$. Dividing by the negative number $-2$ reverses the direction: $x\le -2m-1$.

2. 2

   Translate no solution into a condition on m. The system asks for $x>m+2$ and $x\le -2m-1$ at the same time. Such an $x$ fails to exist exactly when the upper bound does not sit strictly above the lower bound, i.e. $-2m-1\le m+2$. That gives $-3\le 3m$, so $m\ge -1$.

3. 3

   Solve the linear equation. $x+m-2=2-x$ gives $2x=4-m$, so $x=\frac{4-m}{2}$.

4. 4

   Impose non-negative integer on that root. Non-negativity needs $\frac{4-m}{2}\ge 0$, i.e. $m\le 4$. Integrality needs $4-m$ to be even, so $m$ must be even.

5. 5

   Intersect the three conditions. Combining $m\ge -1$, $m\le 4$ and $m$ even over the integers leaves $m\in\{0,2,4\}$.

6. 6

   Check each survivor. $m=0$: system needs $x>2$ and $x\le -1$ (empty), root $x=2$. $m=2$: $x>4$ and $x\le -5$ (empty), root $x=1$. $m=4$: $x>6$ and $x\le -9$ (empty), root $x=0$. All three qualify, and $0+2+4=6$.