Question 1202420: Find four consecutive even integers such that the product of -12 and the sum of the first and fourth is 6 less than the product of 19 and the opposite of the third.
Answer by math_tutor2020(3817)   (Show Source):
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x = first even integer
x+2 = second even integer
x+4 = third even integer
x+6 = fourth even integer
first+fourth = x+(x+6) = 2x+6
product of -12 with that previous expression gives -12(2x+6) = -24x-72
That expression is "6 less than the product of 19 and the opposite of the third", so,
-24x-72 = 19*(-(x+4)) - 6
-24x-72 = -19(x+4) - 6
-24x-72 = -19x-76 - 6
-24x-72 = -19x-82
-24x+19x = -82+72
-5x = -10
x = -10/(-5)
x = 2
Then,
x+2 = 2+2 = 4
x+4 = 2+4 = 6
x+6 = 2+6 = 8
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Answer:
The consecutive even integers are 2,4,6,8
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