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solve the equation in natural numbers as
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usual pause the video before watching
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and try to solve it divide both sides of
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the equation by a which is not equal to
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zero since we are solving in natural
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numbers we get 1 + B / a = b * C now
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divide both sides of the equation by B
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which is also not equal to zero and we
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get a / B + 1 = a * C each fraction a /
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B and B / a is greater than than zero
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and the sum with one gives the product
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of two natural numbers meaning both
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fractions must also be natural numbers
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divided b equals m where M belongs to
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the set of natural numbers and B / a
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must equal n where n also belongs to the
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set of natural numbers from the first
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equation we find a = b * m and we
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substitute this into the second equation
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to get B / by B * m = n simplifying we
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get 1 / m = n from this we find that M *
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n equal 1 and since the product of two
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natural numbers equals 1 both M and N
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must equal 1 substituting this back into
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the first equation we find a equal B and
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considering that a equal B we return to
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our system from the first equation we
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get B * c = 2 and from the second
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equation a * c = 2 this gives us two
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cases to consider the first case is when
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a = b = 1 and C = 2 substituting these
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values into the first equation we get 1
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+ 1 = 1 * 1 * 2 which simplifies 2 2 = 2
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confirming that this is a valid solution
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the second case is when a = b = 2 and a
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c = 1 substituting these values into the
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equation we get 2 + 2 = 2 * 2 * 1 which
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simplifies to 4 = 4 confirming that this
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is also a valid solution therefore the
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solution to the equation consists of two
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sets of numbers 1 1 2 and 2 2 1 the
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problem is solved if you understood the
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solution leave a like write a comment
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