# SOLUTION: Find 3 consecutive multiples of six such that 4 times the first exceeds twice the third by 12.

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 Click here to see ALL problems on real-numbers Question 123513: Find 3 consecutive multiples of six such that 4 times the first exceeds twice the third by 12.Answer by ankor@dixie-net.com(15746)   (Show Source): You can put this solution on YOUR website!Find 3 consecutive multiples of six such that 4 times the first exceeds twice the third by 12. : Let x = 1st multiply of 6 Then (x+6) = 2nd multiply of 6 and (x+12) = 3rd multiply of 6 : Write an equation from the statement, "4 times the first exceeds twice the third by 12." : 4x = 2(x+12) + 12 : 4x = 2x + 24 + 12 : 4x - 2x = 36 : 2x = 36 : x = 18 : The three multiples of 6 are: 18, 24, 30 : Check solution in the statement: "4 times the first exceeds twice the third by 12." 4(18) = 2(30) + 12 72 = 60 + 12