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Carrie
Average Member
USA
17 Posts |
 Posted - 03/04/2009 : 12:43:56
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Here's the problem:
Choose three consecutive odd integers such that three times the second decreased by 4 is equal to twice the third increased by 15.
Here's my stab at it:
Let x = 1st integer
Let x+2 = 2nd integer
Let x+4 = 3rd integer
3(x+2)-4=2(X+4)+15
I solve and come up with x=21, 23.25 which works in my equation for what it's worth.
The answers are 11,13, and 15 which don't work in my equation.
What am I missing?
Any help would be greatly appreciated.
Carrie |
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Mrspi
Advanced Member
USA
994 Posts |
Posted - 03/04/2009 : 15:23:45
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quote:
Originally posted by Carrie
Here's the problem:
Choose three consecutive odd integers such that three times the second decreased by 4 is equal to twice the third increased by 15.
Here's my stab at it:
Let x = 1st integer
Let x+2 = 2nd integer
Let x+4 = 3rd integer
3(x+2)-4=2(X+4)+15
I solve and come up with x=21, 23.25 which works in my equation for what it's worth.
The answers are 11,13, and 15 which don't work in my equation.
What am I missing?
Any help would be greatly appreciated.
Carrie
Your equation and your solution look fine to me.
The numbers 11, 13, and 15 do NOT fit the conditions of the problem:
3 times the second number is 39, and 39 decreased by 4 is 35.
2 times the third number is 30, and 30 increased by 15 is 45.
35 is not equal to 45.....
Perhaps the solutions given in your text are incorrect...it wouldn't be the first time I've known that to happen.
I hope this helps you. |
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Carrie
Average Member
USA
17 Posts |
Posted - 03/04/2009 : 16:41:46
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Thanks. I was wondering. Thanks for your time to confirm my work!
I appreciate it.
Carrie |
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Topic |
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Algebra solving for consecutive odd integers
