6. A collection of coins consisting of dimes and quarters amounts to $5.35. Twice the number of dimes is equal to 5 more than the number of
quarters. How many of each type of coins are there?

11 Dimes

17 Quarters

Step-by-step explanation:

First, you must create an algebraic system out of the situation.

Let D represent the number of dimes and Q represent the number of quarters.

You can make the first equation using the total amount of money all coins equal. Since every dime is 10 cents and every quarter is 25 cents, the equation is:

0.10 D + 0.25 Q = 5.35

The second equation used to represent the ratio between dimes and quarters. It says that  twice the number of dimes is = 5 more than the number of quarters. Replace dimes with D and quarters with Q to get:

2D = Q + 5

The system of equations is:

0.10 D + 0.25 Q = 5.35

2D = Q + 5

Solve for each variable.

Solving for D:

0.10 D + 0.25 Q = 5.35

Q = 2D - 5

Plug in:

0.10 D + 0.25(2D - 5) = 5.35

0.10D + 0.5D - 1.25 = 5.35

0.6D = 5.35 + 1.25 = 6.6

D = 6.6 / 0.6 = 11

Solving for Q:

Q = 2D - 5

Q = 2(11) - 5 = 22 - 5 = 17

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18