A clownfish costs $3.60 to buy. An angelfish costs $5.80 to buy. Joy spends $42.00 on some clownfish and some angelfish. She buys at least one of each. How many of each type does she buy? What if she had $52 to spend on fish? How many of each type did she buy?

  a) 6 angelfish, 2 clownfish

  b) 4 angelfish, 8 clownfish

Step-by-step explanation:

There are at least a couple of different algorithms for finding integer solutions to problems like this. However, it is easiest to describe a trial-and-error solution.

Here, we have defined a function that tells us the number of clownfish we will get if we purchase some number of angelfish. When the function value is an integer, we have found a solution. We compute the value of the function for all reasonable numbers of angelfish.

a) The problem statement tells us the total purchase amount for "c" clownfish and "a" angelfish will be ...

   3.60c +5.80a = 42.00

We are told that we must buy at least 1 of each kind of fish, so the most clownfish we can buy will be ...

  3.60c +5.80×1 = 42.00

  c = (42.00 -5.80)/3.60 = 10 1/18

And the most angelfish we can buy will be ...

  3.60×1 +5.80a = 42.00

  a = (42.00 -3.60)/5.80 = 6 18/29

Since we can buy fewer angelfish, a trial-and-error solution will look a the number of clownfish we can buy for each different purchase of angelfish.

We can write a function, similar to the equation for "c" above, that tells us the number of clownfish for x angelfish:

  f(x) = (42 -5.80x)/3.60

We want to find the value of x that results in an integer number of clownfish. The attached table shows us that purchase of 6 angelfish will allow purchase of 2 clownfish for $42.

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b) Using the same idea, we can repeat the process for a total purchase of $52. The attached table tells us the solution is a purchase of 4 angelfish and 8 clownfish for $52.

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step-by-step explanation:

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m∠a is 90°

m∠b is 40°