Iara has $\mathrm{R} \$ 50.00$ to buy glasses and plates. Each glass costs $\mathrm{R} \$ 2.50$ and each plate, $\mathrm{R} \$ 7.00$. She wants to buy at least 4 plates and 6 glasses. What can she buy?
Let $c$ and $p$ be the number of glasses and plates Iara can buy. Clearly, $c$ and $p$ are integers, and since she wants to buy at least four plates and six glasses, we have $p \geq 4$ and $c \geq 6$. Since each glass costs $\mathrm{R} \$ 2.50$ and each plate costs $\mathrm{R} \$ 7.00$, the cost of $c$ glasses and $p$ plates is $2.5 c + 7 p$. However, Iara only has $\mathrm{R} \$ 50.00$, so $2.5 c + 7 p \leq 50$. Therefore, we need to find two integers $c$ and $p$ that satisfy $p \geq 4, c \geq 6$, and $2.5 c + 7 p \leq 50$.

- If Iara buys four plates, she has $50 - 4 \times 7 = 22$ reais left for the glasses. Since $22 = 8 \times 2.50 + 2$, she can buy eight more glasses (leaving $\mathrm{R} \$ 2.00$).
- If Iara buys five plates, she has $50 - 5 \times 7 = 15$ reais left for the glasses. Since $15 = 6 \times 2.50$, she can buy six more glasses.
- If Iara buys six plates, she has $50 - 6 \times 7 = 8$ reais left for the glasses, which allows her to buy only three glasses, which is not what she wants.

Therefore, Iara can buy four plates and eight glasses, or five plates and six glasses.