1. **State the problem:** Mayerlin has $x$ dimes and $y$ nickels. She has no more than 20 coins combined: $$x + y \leq 20$$ The total value is no less than 1.30 dollars. Since dimes are 10 cents and nickels are 5 cents: $$0.10x + 0.05y \geq 1.30$$ 2. **Rewrite inequalities for graphing:** Coin count inequality: $$x + y \leq 20$$ Value inequality multiply by 20 to clear decimals: $$2x + y \geq 26$$ Or equivalently: $$y \geq 26 - 2x$$ 3. **Explain graph:** - The region satisfying $x + y \leq 20$ is the area below or on the line $y = 20 - x$. - The region satisfying $y \geq 26 - 2x$ is the area above or on the line $y = 26 - 2x$. 4. **Find intersection points of two lines:** Set $20 - x = 26 - 2x$ $$20 - x = 26 - 2x$$ Add $2x$ to both sides: $$20 + x = 26$$ Subtract 20: $$x = 6$$ Find $y$: $$y = 20 - x = 20 - 6 = 14$$ So, intersection is at $(6, 14)$. 5. **Determine possible solution:** Because $x$ and $y$ represent counts of coins, they must be integers and non-negative. Try $x=7$, then $y$ should satisfy: $$x + y \leq 20 \Rightarrow 7 + y \leq 20 \Rightarrow y \leq 13$$ And also $$y \geq 26 - 2(7) = 26 - 14 = 12$$ Therefore, $12 \leq y \leq 13$. Choosing $y=12$: - Total coins: $7 + 12 = 19 \leq 20$ (okay) - Value: $0.10 \times 7 + 0.05 \times 12 = 0.70 + 0.60 = 1.30 \geq 1.30$ (okay) Thus $(7, 12)$ is one possible solution. **Final answer:** One possible integer solution is $x=7$ dimes and $y=12$ nickels.