Reiny
answered
nickels -- x+3
quarters --- 38-x-(x+3) = 35-2x
10x + 5(x+3) + 25(35-2x) = 795
10x + 5x + 15 + 875 - 50x = 795
-35x = -80
no integer solution, so the question is flawed.
check my arithmetic, and check your typing
Explain Bot
answered
Let's assign variables to represent the number of nickels, dimes, and quarters that Gail has. Let:
- N be the number of nickels
- D be the number of dimes
- Q be the number of quarters
According to the problem, Gail has three more nickels than dimes, so we can write the equation: N = D + 3 --- Equation 1
The second piece of information states that the total number of coins Gail has is 38, so the equation is: N + D + Q = 38 --- Equation 2
Next, we need to convert the values of the coins into cents since $7.95 is in cents. A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents.
The value of N nickels in cents is 5N.
The value of D dimes in cents is 10D.
The value of Q quarters in cents is 25Q.
According to the problem, the total value of the coins is $7.95, which is equivalent to 795 cents. So, the equation is: 5N + 10D + 25Q = 795 --- Equation 3
Now we have a system of three equations:
N = D + 3 --- Equation 1
N + D + Q = 38 --- Equation 2
5N + 10D + 25Q = 795 --- Equation 3
To solve this system of equations, we can substitute the value of N from Equation 1 into Equations 2 and 3.
Substituting N = D + 3 in Equation 2, we get: (D + 3) + D + Q = 38
Simplifying, we have: 2D + Q = 35 --- Equation 4 (after rearranging)
Substituting N = D + 3 in Equation 3, we get: 5(D + 3) + 10D + 25Q = 795
Simplifying, we have: 15D + 25Q = 780 --- Equation 5 (after simplifying and rearranging)
Now we have a new system of equations:
2D + Q = 35 --- Equation 4
15D + 25Q = 780 --- Equation 5
Solving this system will give us the values for D (dimes) and Q (quarters). Let's solve it:
From Equation 4, we can express Q in terms of D: Q = 35 - 2D.
Substituting this into Equation 5, we get: 15D + 25(35 - 2D) = 780
Simplifying, we have: 15D + 875 - 50D = 780
Combining like terms, we get: -35D = -95
Dividing both sides by -35, we find: D = 3
Now that we have the value for D, we can substitute it back into Equation 4 to find Q:
2(3) + Q = 35
6 + Q = 35
Subtracting 6 from both sides, we get: Q = 29
Finally, substituting the values of D (3) and Q (29) into Equation 1, we can solve for N:
N = D + 3 = 3 + 3 = 6
So, Gail has 6 nickels, 3 dimes, and 29 quarters.
