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Wes had 11.85 in quarters and dimes. If he had 51 coins, how many of each did he have?

User Jowett
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1 Answer

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answer:

Let's assume that Wes has x number of quarters and y number of dimes. We can set up two equations to represent the given information.

Equation 1: The total value of quarters and dimes is $11.85.

0.25x + 0.10y = 11.85

Equation 2: The total number of coins is 51.

x + y = 51

To solve this system of equations, we can use the method of substitution or elimination. In this case, let's solve it using the elimination method.

First, let's multiply Equation 2 by -0.10 to make the coefficient of y in Equation 2 equal to -0.10y:

-0.10(x + y) = -0.10(51)

-0.10x - 0.10y = -5.10

Now, we can add Equation 1 and Equation 2:

(0.25x + 0.10y) + (-0.10x - 0.10y) = 11.85 - 5.10

0.15x = 6.75

Dividing both sides of the equation by 0.15:

x = 6.75 / 0.15

x = 45

Now, substitute the value of x back into Equation 2 to find the value of y:

45 + y = 51

y = 51 - 45

y = 6

Therefore, Wes has 45 quarters and 6 dimes.

In summary:

- Number of quarters (x): 45

- Number of dimes (y): 6

Alli <3

User Andreas Schwarz
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