Question 180404This question is from textbook Beginning Algebra
: i can't work this out. the answer, according to the book, is 15:
if 6 is subtracted from the third of three consecutive odd integers, and the result is multiplied by 2, the answer is 23 less than the sum of the first and two times the second of the three integers.
x-4-6= x-2 x 2= 2x-4
x + 2(x +2)-23
2x-4 = x + 2(x+2) + 23
the book says the answe is 15. i don't get anywhere near that answer. thanks!!
This question is from textbook Beginning Algebra
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! i can't work this out. the answer, according to the book, is 15:
if 6 is subtracted from the third of three consecutive odd integers, and the result is multiplied by 2, the answer is 23 less than the sum of the first and two times the second of the three integers.
x-4-6= x-2 x 2= 2x-4
x + 2(x +2)-23
2x-4 = x + 2(x+2) + 23
the book says the answe is 15. i don't get anywhere near that answer. thanks!!
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If x is the first, or smallest, integer, the other 2 are x+2 and x+4.
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if 6 is subtracted from the third of three consecutive odd integers --> x+4 - 6
That's x-2
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and the result is multiplied by 2
That's 2x-4
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the answer is 23 less than the sum of the first and two times the second of the three integers
The sum of the 1st and 2 times the 2nd is x + 2*(x+2) = 3x+4
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If we add 23 to the 2x-4, then they'll be equal.
2x-4 + 23 = 3x+4
2x+19 = 3x+4
x = 15
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You went astray somewhere, subtracted when you should have added, maybe.
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