Question 625488
"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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