Question 625488: Hi my problem says: find three consecutive odd integers such that 7 times the sum of the first and the third is 120 less than 10 times the OPPOSITE of the second.
Please show work :) thank you
Found 2 solutions by oscargut, jim_thompson5910:
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website! Those 3 consecutive odd integers are: n, n+2 and n+4
7(n+n+4) = 10(-n-2)-120
7(2n+4) = -10n-20-120
14n+28 = -10n -140
14n + 10n = -140 -28
24n = -168
n = -168/24
n = -7
Answer: -7, -5 and -3
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Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! "three consecutive odd integers such that 7 times the sum of the first and the third is 120 less than 10 times the OPPOSITE of the second" translates to
7*( (2x+1) + (2x+5) ) = 10(-1(2x+3)) - 120
This is because 2x+1 is an odd integer, 2x+3 is the next consecutive odd integer, and 2x+5 is the next consecutive odd integer after that
Let's solve for x
7*( (2x+1) + (2x+5) ) = 10(-1(2x+3)) - 120
7*( (2x+1) + (2x+5) ) = -10(2x+3) - 120
7*( 2x+1 + 2x+5 ) = -10(2x+3) - 120
7*( 4x+6 ) = -10(2x+3) - 120
28x+42 = -20x - 30 - 120
28x+42 = -20x - 150
28x+20x = -150-42
48x = -192
x = -192/48
x = -4
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Now that we know that x = -4, we can plug it into 2x+1, 2x+3, and 2x+5 to find the three consecutive odd integers
2x+1 = 2(-4)+1 = -7
2x+3 = 2(-4)+3 = -5
2x+5 = 2(-4)+5 = -3
So the three consecutive odd integers are -7, -5, and -3
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