Danielle and Emilia brought apples, bananas, and oranges to a fruit sale. The bananas were sold for $0.50 each, while the apples and oranges
were sold for $0.75 each. They sold 50 pieces of fruit and earned $33.50
total. If Danielle and Emilia sold twice as many bananas as oranges, how
many apples did they sell? Show your work.
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Can you please Write the system which models this situation as well!!

26 apples

Step-by-step explanation:

bananas, b= $0.50

oranges, o= $0.75

apples, a = $0.75  

Total fruit sold = 50

Total Earned  = $33.50

b + o + a = 50       Because the total number of all 3 fruits sold is 50

.5b + .75o + .75a =  $33.50           The number of each fruit, multiplied by the price of each fruit, and added together is $33.50

They sold twice as many bananas than oranges.

If o is oranges, then 2o = b because there are twice as many bananas so:

b + o + a = 50

2o + o + a = 50  

3o + a = 50

Now we can solve.

3o + a = 50

a = 50 - 3o

.5b + .75o + .75a =  $33.50

.5(2o) + .75o + .75a = $33.50

o + .75o + .75a = $33.50

1.75o + .75a = $33.50

Two equations and 2 unknowns.

a = 50 - 3o

1.75o + .75a = $33.50       Plug in a to solve for o.

1.75o + .75(50 - 3o) = $33.50

1.75o + .75*50 - .75*3o = $33.50

1.75o + 37.5 - 2.25o = $33.50   Combine like terms

1.75o - 2.25o +37.5 = $33.50

-.5o + 37.5 = 33.5     Add .5o to each side so it cancels on the left.

37.5 = 33.5 + .5o   Subtract 33.5 from each side

37.5 -33.5 = .5o

4 = .5o   Divide each side by .5

4/.5 = o

8 = o

Now plug into a = 50 - 3o

a = 50 - 3(8)

a = 50 - 24

a = 26

So even though we know they sold 26 apples, let's finish solving to check our work.

b + o + a = 50

b + 8 + 26 = 50

b + 34 = 50

b = 50 - 34

b = 16      

b = 2o  

b = 2 (8)

b = 16   Which works! Now the last step of checking our work:

.5b + .75o + .75a =  $33.50

.5(16) + .75(8) + .75(26) =  $33.50

8 + 6 + 19.5 = $33.50

14 + 19.5 = $33.50

33.5 = $33.50  Yes! It works too!

bananas = 16

oranges = 8

apples = 26