Answer:
To solve this problem algebraically, we can set up a system of equations based on the given information.
Let's represent the number of dimes as 'd' and the number of quarters as 'q'. (To make it more simple)
From the information provided, we can establish two equations:
The total value equation: 0.10d + 0.25q = 18.50
The total number of coins equation: d + q = 110
We can now solve this system of equations using the elimination method.
First, we'll multiply the second equation by 0.10 to align the coefficients of 'd' in both equations:
0.10d + 0.10q = 11.00
Now, we can subtract this equation from the first equation to eliminate 'd':
(0.10d + 0.25q) - (0.10d + 0.10q) = 18.50 - 11.00
This simplifies to:
0.15q = 7.50
Next, we divide both sides by 0.15:
q = 7.50 / 0.15
q = 50
Now that we know q = 50, we can substitute this value back into the second equation to solve for 'd':
d + 50 = 110
Subtracting 50 from both sides:
d = 110 - 50
d = 60
Therefore, there are 60 dimes and 50 quarters in the jar.
In conclusion, there are 60 dimes and 50 quarters in the jar, based on the given information and solving the system of equations algebraically.