1. **State the problem:** We have pennies (1 cent), nickels (5 cents), and dimes (10 cents). There are 13 coins total, worth 83 cents. We want to find how many of each coin there are. 2. **Define variables:** Let $x$ = number of pennies, $y$ = number of nickels, $z$ = number of dimes. 3. **Write equations:** - Total coins: $$x + y + z = 13$$ - Total value: $$1x + 5y + 10z = 83$$ 4. **Set up matrix form:** $$\begin{bmatrix}1 & 1 & 1 \\ 1 & 5 & 10\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}13 \\ 83\end{bmatrix}$$ 5. **Since we have 2 equations and 3 unknowns, express $x$ in terms of $y$ and $z$ from the first equation:** $$x = 13 - y - z$$ 6. **Substitute into the value equation:** $$1(13 - y - z) + 5y + 10z = 83$$ 7. **Simplify:** $$13 - y - z + 5y + 10z = 83$$ $$13 + 4y + 9z = 83$$ 8. **Subtract 13 from both sides:** $$4y + 9z = 70$$ 9. **Find integer solutions for $y$ and $z$ that satisfy $4y + 9z = 70$ with $y,z \geq 0$ and $x = 13 - y - z \geq 0$:** Try values of $z$: - For $z=2$: $4y + 18 = 70 \Rightarrow 4y = 52 \Rightarrow y = 13$ (too large, $x=13-13-2=-2$ no) - For $z=4$: $4y + 36 = 70 \Rightarrow 4y = 34 \Rightarrow y = 8.5$ (not integer) - For $z=5$: $4y + 45 = 70 \Rightarrow 4y = 25 \Rightarrow y = 6.25$ (not integer) - For $z=7$: $4y + 63 = 70 \Rightarrow 4y = 7 \Rightarrow y = 1.75$ (not integer) - For $z=10$: $4y + 90 = 70$ no (negative) Try $z=3$: $$4y + 27 = 70 \Rightarrow 4y = 43 \Rightarrow y = 10.75$$ no Try $z=1$: $$4y + 9 = 70 \Rightarrow 4y = 61 \Rightarrow y = 15.25$$ no Try $z=0$: $$4y = 70 \Rightarrow y = 17.5$$ no Try $z=6$: $$4y + 54 = 70 \Rightarrow 4y = 16 \Rightarrow y = 4$$ integer! Check $x$: $$x = 13 - 4 - 6 = 3$$ valid. 10. **Solution:** $$x=3$$ pennies, $$y=4$$ nickels, $$z=6$$ dimes. 11. **Verify total value:** $$3 \times 1 + 4 \times 5 + 6 \times 10 = 3 + 20 + 60 = 83$$ cents correct. **Final answer:** There are 3 pennies, 4 nickels, and 6 dimes in the box.