SOLUTION: Find three consecutive integers such that the sum of the second and third exceeds half of the first by 33. This is what I tried.... (x+1)+(x+2)=1/2x+33 which gave me x=30

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find three consecutive integers such that the sum of the second and third exceeds half of the first by 33. This is what I tried.... (x+1)+(x+2)=1/2x+33 which gave me x=30      Log On


   

Question 202675: Find three consecutive integers such that the sum of the second and third exceeds half of the first by 33.
This is what I tried.... (x+1)+(x+2)=1/2x+33 which gave me x=30
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B1%29%2B%28x%2B2%29=%281%2F2%29x%2B33 Start with the given equation.


2x%2B3=%281%2F2%29x%2B33 Combine like terms.


2%282x%29%2B2%283%29=cross%282%29%28%281%2Fcross%282%29%29x%29%2B2%2833%29 Multiply EVERY term by the LCD 2 to clear out the fraction.


4x%2B6=x%2B66 Multiply and simplify.


4x=x%2B66-6 Subtract 6 from both sides.


4x-x=66-6 Subtract x from both sides.


3x=66-6 Combine like terms on the left side.


3x=60 Combine like terms on the right side.


x=%2860%29%2F%283%29 Divide both sides by 3 to isolate x.


x=20 Reduce.


----------------------------------------------------------------------

Answer:

So the solution is x=20


This means that the first number is 20, the second is 21 (20+1=21), and the third is 22 (20+2=22)