1. **State the problem:** A coin collector has 42 coins in total, some are 10-value coins and the rest are 20-value coins. The total face value of all coins is 650. We need to find how many coins of each denomination the collector has. 2. **Define variables:** Let $x$ be the number of 10-value coins and $y$ be the number of 20-value coins. 3. **Write the system of equations:** $$\begin{cases} x + y = 42 \\ 10x + 20y = 650 \end{cases}$$ 4. **Solve the first equation for $y$:** $$y = 42 - x$$ 5. **Substitute $y$ into the second equation:** $$10x + 20(42 - x) = 650$$ 6. **Distribute 20:** $$10x + 840 - 20x = 650$$ 7. **Combine like terms:** $$\cancel{10x} - 20x + 840 = 650$$ $$-10x + 840 = 650$$ 8. **Subtract 840 from both sides:** $$-10x = 650 - 840$$ $$-10x = -190$$ 9. **Divide both sides by -10:** $$\frac{-10x}{\cancel{-10}} = \frac{-190}{\cancel{-10}}$$ $$x = 19$$ 10. **Find $y$ using $y = 42 - x$:** $$y = 42 - 19 = 23$$ **Final answer:** The collector has 19 coins of 10-value and 23 coins of 20-value.