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emmainna [20.7K]
2 years ago
7

Determine the number of solution pairs in the equation.

Mathematics
2 answers:
kap26 [50]2 years ago
7 0

\\ \tt\longmapsto 2n=3k

\\ \tt\longmapsto n\propto k

  • Both n and k are positive i.e n,k>0

So the domain of n is

\\ \tt\longmapsto n\in [0,100]

Now

  • N has total 100+1=101 integer values
  • K will also have 100+1=101values .

So total solution sets are 101 .

chubhunter [2.5K]2 years ago
7 0

Answer:

  • 33 solutions

Step-by-step explanation:

<u>From the relationship we can see:</u>

  • 2n = 3k
  • n = 3/2k
  • n = 1.5k

<u>Since we are looking for integer solution for both variables, any value of k = 2p will give us:</u>

  • n = 3p

It means n can be a multiple of 3.

<u>Multiples of 3 are:</u>

  • 3, 6, ... , 99

The number of multiples of 3 not greater than 100 is 33 and for any n we have matching integer k.

Since no restrictions on the value of k, apart from this being an integer, the number of solution pairs is same as the number of values of n = 33.

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Answers straight from plato/Edmentum Hope this helps:) !

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