lavre has sixty-one coins, all of whichare dimes and quaters. if the total value of the coins is $9.85, how many of each kind of coins has she?
Q = 25
D = 36
It's a system of linear equations so you can put it into a matrix to solve
let Matrix A be a 2x2 matrix
|1 1|
|.25 .1|
and have another matrix, B
|61 |
|9.85|
A^-1 * B will give you the answer
To solve this problem, we can use a system of equations. Let's assume that Lavre has x dimes and y quarters.
Since Lavre has sixty-one coins in total, we can write the first equation as:
x + y = 61 (Equation 1)
We also know that the total value of all the coins is $9.85. We can find the value of the dimes and quarters, and then equate it to the total value.
The value of x dimes is 10x cents, and the value of y quarters is 25y cents. So the second equation can be written as:
10x + 25y = 985 (Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) that we can solve to find the values of x and y.
We can use substitution or elimination method to solve this system of equations.
Let's use the elimination method to solve it:
From Equation 1, we know that x + y = 61. Let's multiply this equation by 10 to make the coefficients of x and y equal:
10x + 10y = 610 (Equation 3)
Now, subtracting Equation 3 from Equation 2, we get:
(10x + 25y) - (10x + 10y) = 985 - 610
15y = 375
Dividing both sides of the equation by 15, we find that:
y = 25
Substituting the value of y back into Equation 1, we get:
x + 25 = 61
x = 61 - 25
x = 36
Therefore, Lavre has 36 dimes and 25 quarters.
If Q = # quarters
and D = # dimes.
Then
Q+D=61
.25Q+.10D=9.85
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Two equations and two unknowns. Solve for Q and D. Check my work.