Question 499628: How can I find 3 consecutive integers whose sum is 33?
This is what I have tried.
x+(x+1)+(x+2)=33
I combined the like terms next and got 3x+3=33
I then subtracted three from both sides. -3 -3
I got the answer of 3x=33
after that I divided by 3 on both sides 3x/3 33/3
I got the answer of X=11
My 3 consecutive integers are 11,12,and 13.
Is my answer correct?
Thank you!
Found 2 solutions by Alan3354, Earlsdon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! How can I find 3 consecutive integers whose sum is 33?
This is what I have tried.
x+(x+1)+(x+2)=33
I combined the like terms next and got 3x+3=33
I then subtracted three from both sides. -3 -3
I got the answer of 3x=33 ************ 33 - 3 = 30, not 33
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after that I divided by 3 on both sides 3x/3 33/3
I got the answer of X=11
My 3 consecutive integers are 11,12,and 13.
Is my answer correct?
------------
Add them and see.
11 + 12 + 13 = 36 <> 33 so no.
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I combined the like terms next and got 3x+3=33
I then subtracted three from both sides. -3 -3
3x = 30
x = 10
--> 10, 11 & 12
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! No, it isn't!
When you subtracted 3 from both sides of:
3x+3 = 33 you should have got:
3x = 30 not 3x = 33, so...
3x = 30 Divide by 3 to get:
x = 10
The three consecutive numbers ar:
10, 11, and 12 whose sum is 33.
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