# The sq of a nat num is 4 more than 2 times another nat num. The 2nd num is 2 less than 3 times the 1st num. ? the 2 nat num algebraically.?

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Let x be the first natural number.

Let y be the second natural number.

x^2 = 2y + 4

y = 3x - 2

Substituting gives:

x^2 = 2(3x - 2) + 4

x^2 = 6x - 4 + 4

x^2 = 6x

x^2 - 6x = 0

x(x - 6) = 0

x = 6 is a solution. (x = 0 is not a natural number, 1,2,3...)

y = 3x - 2 = 3(6) - 2 = 18 - 2 = 16.

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• "The sq of a nat num is 4 more than 2 times another nat num."

m is the first number; n is the second number.

m² = 2n+4

"The 2nd num is 2 less than 3 times the 1st num."

n = 3m-2

m² = 2(3m-2) + 4

m² = 6m

m = 6

n = 3m-2 = 16

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• The square of a natural number is 4 more than 2 times another natural number:

x² = 2y + 4

The second number is 2 less than 3 times the first number:

y = 3x - 2

we have a system of two equations and two unknowns.  Substitute and solve:

x² = 2y + 4

x² = 2(3x - 2) + 4

x² = 6x - 4 + 4

x² = 6x

x² - 6x = 0

x(x - 6) = 0

x = 0 and 6

Since we are told that these are natural numbers, we can throw out the zero, leaving:

x = 6

Now we can solve for y to determine the second number:

y = 3x - 2

y = 3(6) - 2

y = 18 - 2

y = 16

The second number is 16

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• M^2 = 2N + 4 and

N + 2 = 3M.

Multiplying the 2nd eqn by 2, we see that

M^2 = 6M, so M = 6 and therefore N = 16.

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• x² = 2y + 4

y = 3x – 2

x² = 2(3x – 2) + 4

x² = 6x – 4 + 4

x² = 6x

x² – 6x = 0

x(x – 6) = 0

x = 0, 6

y = 3x – 2

y = 3•0 – 2 = –2 (this is ruled out as negative numbers are not allowed)

y = 3•6 – 2 = 16

ans: 6 and 16

Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0, corresponding to the non-negative integers 0, 1, 2, 3, …, whereas others start with 1, corresponding to the positive integers 1, 2, 3, …, while others acknowledge both definitions.

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