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Question 414691: The second of three numbers is 8 more than twice the first. The third number is twelve less than the first. The sum of the number is 108. Find the numbers
work so far--> 2n+8+n-12=108
3n+8-12=108 added the n's
3n+8=120 added 12
3n=112 subtracted 8
when I divided 112/3 I get an irrational number and dont know what to do after
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! The second of three numbers is 8 more than twice the first. The third number is twelve less than the first. The sum of the number is 108. Find the numbers
work so far--> 2n+8+n-12=108<-----------------*you forgot to add "n" the 1st number*
3n+8-12=108 added the n's
3n+8=120 added 12
3n=112 subtracted 8
when I divided 112/3 I get an irrational number and dont know what to do after
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n = first number
2n + 8 = second number {8 more than twice the first}
n - 12 = third number {12 less than the first}
n + 2n + 8 + n - 12 = 108 {sum of numbers is 108}
4n - 4 = 108 {combined like terms}
4n = 112 {added 4 to both sides}
n = 28 {divided both sides by 4}
2n + 8 = 64 {substituted 28, in for n, into 2n + 8}
n - 12 = 16 {substituted 28, in for n, into n - 12}
28, 64, and 16 are the three numbers
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