Asked by Bob
three numbers are such that the second is the difference of three times the first and 6 while the third is the sum of 2 and 2/3 the second. The sum of the three numbers is 172. Find the largest number.
Answers
Answered by
Anonymous
n1= first number
n2 = second number
n3 = third number
n2 = 3 n1 - 6
n3 = 2 + ( 2 / 3 ) n2
n3 = 2 + ( 2 / 3 ) * ( 3 n1 - 6 )
n3 = 2 + ( 2 / 3 ) * 3 n1 - ( 2 / 3 ) * 6
n3 = 2 + 2 n1 - 4
n3 = 2 n1 - 2
n1 + n2 + n3 = 172
n1 + 3 n1 - 6 + 2 n1 - 2 = 172
6 n1 - 8 = 172 add 8 to both sides
6 n1 - 8 + 8 = 172 + 8
6 n1 = 180 Divide both sides by 6
n1 = 180 / 6 = 30
n2 = 3 n1 - 6
n2 = 3 * 30 - 6 = 90 - 6 = 84
n3 = 2 n1 - 2
n3 = 2 * 30 - 2 = 60 - 2 = 58
The largest number = n2 = 84
n2 = second number
n3 = third number
n2 = 3 n1 - 6
n3 = 2 + ( 2 / 3 ) n2
n3 = 2 + ( 2 / 3 ) * ( 3 n1 - 6 )
n3 = 2 + ( 2 / 3 ) * 3 n1 - ( 2 / 3 ) * 6
n3 = 2 + 2 n1 - 4
n3 = 2 n1 - 2
n1 + n2 + n3 = 172
n1 + 3 n1 - 6 + 2 n1 - 2 = 172
6 n1 - 8 = 172 add 8 to both sides
6 n1 - 8 + 8 = 172 + 8
6 n1 = 180 Divide both sides by 6
n1 = 180 / 6 = 30
n2 = 3 n1 - 6
n2 = 3 * 30 - 6 = 90 - 6 = 84
n3 = 2 n1 - 2
n3 = 2 * 30 - 2 = 60 - 2 = 58
The largest number = n2 = 84
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