1. **State the problem:** Kelly and Sarah need at least 12 dollars in change using quarters and dimes. Let $x$ be the number of quarters and $y$ be the number of dimes. 2. **Translate the money requirement into an inequality:** Each quarter is worth 25 cents and each dime 10 cents. The total value must be at least 12 dollars, or 1200 cents. $$25x + 10y \geq 1200$$ 3. **Simplify the inequality by dividing all terms by 5:** $$\frac{25x}{5} + \frac{10y}{5} \geq \frac{1200}{5}$$ $$5x + 2y \geq 240$$ 4. **Express $y$ in terms of $x$ to get a linear inequality:** $$2y \geq 240 - 5x$$ $$y \geq \frac{240 - 5x}{2}$$ $$y \geq 120 - 2.5x$$ 5. **Translate the quarters and dimes count requirement:** They need at least six more quarters than dimes. $$x \geq y + 6$$ 6. **Rewrite this inequality in terms of $y$:** $$y \leq x - 6$$ 7. **Combine the inequalities to form the system:** $$y \geq 120 - 2.5x$$ $$y \leq x - 6$$ 8. **Choose the correct option from the given choices:** The system is: $$y \geq -2.5x + 120, \quad y \leq x - 6$$ Hence, the correct choice is: "y >= -2.5x + 120, y < x - 6" (noting the strict inequality in the problem is "at least", so $\leq$ is appropriate for the second inequality, but the closest choice is with $y < x - 6$).