You have a total of 52 dimes and quarters. You have 10 more quarters than dimes. Which system of equations can you use to find the number z of quarters and number y of dimes you have? Use the system to
determine how much money you have in quarters and dimes.

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  z + y = 52; z - y = 10; z = 31; y = 21; $9.85 total

Step-by-step explanation:

The variables are defined in the problem statement. The equations could be ...

  z + y = 52 . . . . . . total number of coins

  z - y = 10 . . . . . . . 10 more quarters than dimes

These equations can be solved by adding them together.

  (z +y) +(z -y) = (52) + (10)

  2z = 62 . . . . . . . simplify

  z = 31 . . . . . . . . . divide by 2

  31 -10 = y = 21 . . . . find the number of dimes

There are 31 quarters and 21 dimes.

The value of the coins is ...

  31($0.25) +21($0.10) = $7.75 +2.10 = $9.85

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.

any options?

step-by-step explanation: