1. **State the problem:** Daniel bought three distinct types of toys with a total cost of 25. One toy costs 20 more than three times the sum of the other two toys. The total cost of the three toys multiplied by 5 equals 125. 2. **Define variables:** Let the costs of the three toys be $x$, $y$, and $z$. 3. **Write the system of linear equations:** - Total cost: $$x + y + z = 25$$ - One toy costs 20 more than three times the sum of the other two: $$z = 3(x + y) + 20$$ - Total cost multiplied by 5 equals 125: $$5(x + y + z) = 125$$ 4. **Simplify the third equation:** $$5(x + y + z) = 125 \implies x + y + z = \frac{125}{5} = 25$$ This is the same as the first equation, so it does not add new information. 5. **Substitute $z$ from the second equation into the first:** $$x + y + 3(x + y) + 20 = 25$$ $$x + y + 3x + 3y + 20 = 25$$ $$4x + 4y + 20 = 25$$ 6. **Simplify:** $$4x + 4y = 25 - 20 = 5$$ Divide both sides by 4: $$\cancel{4}x + \cancel{4}y = \frac{5}{\cancel{4}} \implies x + y = \frac{5}{4}$$ 7. **Find $z$ using $z = 3(x + y) + 20$:** $$z = 3 \times \frac{5}{4} + 20 = \frac{15}{4} + 20 = \frac{15}{4} + \frac{80}{4} = \frac{95}{4}$$ 8. **Summary:** - $x + y = \frac{5}{4}$ - $z = \frac{95}{4}$ 9. **Interpretation:** The system has infinitely many solutions for $x$ and $y$ satisfying $x + y = \frac{5}{4}$, and $z$ is fixed at $\frac{95}{4}$. The system is consistent and dependent. 10. **Example solution:** If $x = 1$, then $y = \frac{5}{4} - 1 = \frac{1}{4}$. So the costs could be $x=1$, $y=\frac{1}{4}$, $z=\frac{95}{4}$. **Final answer:** - System of equations: $$\begin{cases} x + y + z = 25 \\ z = 3(x + y) + 20 \\ 5(x + y + z) = 125 \end{cases}$$ - Nature: Consistent and dependent with infinitely many solutions. - Costs: $x$ and $y$ satisfy $x + y = \frac{5}{4}$, $z = \frac{95}{4}$.