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Question 896475: when the sum of two whole numbers is multiplied by the difference of the number, the result is 85. If the difference of two numbers is not 1, what is their sum? what are the numbers?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the solution is, i believe, a = 11 and b = 6
you get (a+b)*(a-b) = (11+6) * (11-6) = 17 * 5 = 85
i solved it graphically.
i couldn't see any other way except to create a table and then populate the table until you got a number for a that was a whole number.
there are lots of values, but only 2 values that i could find that were whole numbers.
those values were a = 11 and b = 6
the other values were a = 43 and b = 42
since the difference could not be equal to 1, the only other number i could find was a = 11 and b = 6.
since (a+b)*(a-b) = a^2 - b^2, i solved for a^2 to get a^2 = b^2 + 85
what was left then was to find a number of b^2 + 85 that was a perfect square.
perfect squares greater than 85 were
100
121
144
etc
subtracting 85 from them needs to yield a perfect square as well.
100 - 85 = 15 which is not a perfect square.
121 - 85 = 36 which is a perfect square.
144 - 85 = 59 which is not a perfect square.
36 was the one.
that led to b = 6
that also led to a = 11 because sqrt(121) = 11 and sqrt(36+85) = 11
i then got a^2 - b^2 = 85 becoming 121 - 36 = 85 becoming 85 = 85 confirming that the solution was correct.
graphing was the cleanest and easiest way to find the answer, but you needed to be able to see that the solution was a hole number.
creating an automated table helped with that.
it clearly showed that when x = 6, y = 11
the equation that was graphed was y = sqrt(x^2 + 85)
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