1. **State the problem:** We have a quadratic function $y = x^2 + ax + b$ where $a,b \in \mathbb{Z}$. Points $P(-1,-2)$ and $Q(3,2)$ lie on the curve. We need to find two equations involving $a$ and $b$ using these points. 2. **Use the points to form equations:** Since $P(-1,-2)$ is on the curve, substitute $x=-1$ and $y=-2$ into the equation: $$-2 = (-1)^2 + a(-1) + b = 1 - a + b$$ Rearranged: $$-2 = 1 - a + b$$ 3. **Simplify the first equation:** $$-2 - 1 = -a + b \implies -3 = -a + b$$ Rewrite: $$-a + b = -3$$ 4. **Use point $Q(3,2)$:** Substitute $x=3$ and $y=2$: $$2 = 3^2 + 3a + b = 9 + 3a + b$$ 5. **Simplify the second equation:** $$2 - 9 = 3a + b \implies -7 = 3a + b$$ Rewrite: $$3a + b = -7$$ 6. **Final system of equations:** $$\begin{cases} -a + b = -3 \\ 3a + b = -7 \end{cases}$$ These two equations can be used to solve for $a$ and $b$. **Answer:** The system formed is: $$\boxed{\begin{cases} -a + b = -3 \\ 3a + b = -7 \end{cases}}$$