A restaurant bill of $2.85 was paid using quarters and dimes only. If twice the number of quarters is 6 less than 4 times the number of dimes, how many quarters were used?

# quarters: x

value of quarters: 25x
# dimes: d
value of dimes: 10d
and, 2q = 4d-6, so
q = 2d-3
10d+25q = 285
Now just solve for q

HELP MEEEEEEEEEEEEEEEEEEEEE

Let's solve this problem step-by-step.

Let's assume the number of quarters used as 'q' and the number of dimes used as 'd'.

We know that the value of a quarter is $0.25 and the value of a dime is $0.10.

The total amount paid using quarters and dimes is $2.85.

So we can write the equation:
0.25q + 0.10d = 2.85 ----(1)

We are given that twice the number of quarters is 6 less than 4 times the number of dimes.

Mathematically, this can be written as:
2q = 4d - 6 ----(2)

Now, we have two equations (equation 1 and equation 2) with two variables (q and d). We can solve these equations simultaneously to find the values of q and d.

Let's solve the equations:

From equation (2), we can rewrite it as:
2q - 4d = -6 (multiplying both sides of the equation by -1)

Multiplying equation (1) by 100 to eliminate decimals, we get:
25q + 10d = 285 (multiplying both sides of the equation by 100)

Now we have the system of equations:
2q - 4d = -6 ----(3)
25q + 10d = 285 ----(4)

We can solve this system of equations using either the substitution method or the elimination method.

Let's solve it using the elimination method:

Multiplying equation (3) by 5 and equation (4) by 2 to eliminate the variable 'd', we get:
10q - 20d = -30 ----(5)
50q + 20d = 570 ----(6)

Now, add equation (5) and equation (6) to eliminate 'd':
(10q - 20d) + (50q + 20d) = -30 + 570
60q = 540
Divide both sides of the equation by 60:
q = 540/60
q = 9

Therefore, the number of quarters used is 9.

So, 9 quarters were used.

To solve this problem, let's first set up the equation based on the given information.

Let's assume the number of quarters used is represented by 'q', and the number of dimes used is represented by 'd'.

Given:
1 quarter is equivalent to 0.25 dollars.
1 dime is equivalent to 0.10 dollars.

The total amount paid can be represented as an equation:
0.25q + 0.10d = 2.85

Now, let's translate the second part of the given information into an equation. "Twice the number of quarters is 6 less than 4 times the number of dimes" can be written as:
2q = 4d - 6

Now, we have a system of simultaneous equations:
0.25q + 0.10d = 2.85
2q = 4d - 6

To solve this system of equations, we can use substitution or elimination method. Let's use the elimination method.

Multiply the second equation by 0.25 to make the coefficients of 'q' equal:
0.25(2q) = 0.25(4d - 6)
0.50q = d - 1.50

Now, we have:
0.25q + 0.10d = 2.85
0.50q = d - 1.50

To eliminate the variable 'd' in the second equation, multiply both sides by 0.10:
0.10(0.50q) = 0.10(d - 1.50)
0.05q = 0.10d - 0.15

Now, we have:
0.25q + 0.10d = 2.85
0.05q = 0.10d - 0.15

By multiplying the second equation by 2, we get:
0.10q = 0.20d - 0.30

Let's simplify the two equations:
0.25q + 0.10d = 2.85
0.10q = 0.20d - 0.30

Multiply the second equation by 0.10 to eliminate the decimals:
0.025q + 0.10d = 0.285
0.10q = 0.20d - 0.30

Now, we can subtract the two equations to eliminate 'q':
(0.025q + 0.10d) - (0.25q + 0.10d) = 0.285 - 2.85
-0.225q = -2.565

Divide both sides by -0.225 to solve for 'q':
q = (-2.565) / (-0.225)
q ≈ 11.4

Since we cannot have fractional coins, the number of quarters used is 11.

Therefore, 11 quarters were used to pay the restaurant bill.