*math* How would you work this problem?

Find 3 consecutive integers whose sum is 171





Really need the help-thanks*math* How would you work this problem?
171=a+(a+1)+(a+2)


171=3a+3


174=3a


58=a


58,57,56*math* How would you work this problem?
Think of any 3 consecutive integers:


for example 1, 3, 5 the sum of these is 1+3+5=9





well, the first number is 1


the second number we got by adding two to the first 1+2=3


the third we got by adding four to the first: 1+4=5





but for your problem we dont know which number to start with so we call it n for now. To get the second number we add two to n.


for the third we add 4 to n





1st: n


2nd: n+2


3rd: n+4





n+(n+2)+(n+4)=171 Equation


3n+6=171 Combine like terms


3n=165 Subtract 6 from both sides


n=55 Divide by 3 to both sides





1st:55


2nd:55+2=57


3rd:55+4=59





Solution: 55,57,59
56, 57, 58.





They're consecutive integers, and they add up to 171.
x+x+1+x+2=171


group like things together...


3x+3=171


subtract 3 from each side...


3x=168


divide each side by 3...


x=56


56+57+58=171
x+(x+1)+(x+2)=171


3x=168


x=56


the numbers are 56,57, and 58
x is 1st integer.


x+(x+1)+(x+2)=171


3x+3=171


56,57,58
56+57+58=171