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arlik [135]
3 years ago
15

Find three consecutive positive integers such that the product of the first and​ third, minus the​ second, is 1 more than 7 time

s the third.
Mathematics
1 answer:
Bas_tet [7]3 years ago
6 0
--------------------------------------------------------------------
Define x :
--------------------------------------------------------------------
Let the smallest number be x.
First number = x
Second Number = x + 1
Third Number = x + 2

--------------------------------------------------------------------
Construct equation :
--------------------------------------------------------------------
x(x+2) - (x+1) = 7(x+2) + 1

--------------------------------------------------------------------
Solve x :
--------------------------------------------------------------------
x(x + 2) - (x + 1) = 7(x + 2) + 1
x² + 2x - x - 1 = 7x + 14 + 1
x²- 6x - 16 = 0
<span>(x+2)(x-8) = 0
</span>x = -2 or x = 8

Since x is a positive integer, it cannot be negative.
⇒x = 8

--------------------------------------------------------------------------
Answer: The three numbers are 8, 9 and 10.
--------------------------------------------------------------------------
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Which student correctly used guess and check to determine the unknown number?
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<h2>Explanation:</h2><h2></h2>

Hello! Remember you need to write complete questions in order to get good and exact answers. Here we know nothing about the students so I can only help you providing the correct steps to find <em>the product of a number and eight is 104:</em>

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Let that number be n, then the statement can be written in a mathematical language as follows:

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