SOLUTION: Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!!

Question 31106This question is from textbook Introductory Algebra
: Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!! This question is from textbook Introductory Algebra

Found 2 solutions by blubunny01, venugopalramana:
Answer by blubunny01(20) (Show Source): You can put this solution on YOUR website!
Let x = first integer.
Let y = second integer.
Let z = third integer.

Since they are consecutive integers, we can say the following:
y = x+1 (the next integer after x)
z = y+1 = x+2 (the next integer after y, which is two integers after x)

Then, we can put "first plus one-half the second plus seven less than twice the third is 2101" into an equation:

To solve this equation, we must only have one variable, so we can substitute y and z for the equations we created in terms of x:

Now, it's a matter of simplifying and solving for x:

So, the integers are 601, 602, 603.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!!
LET THE 3 CONSEUTIVE INTEGERS BE X,X+1,X+2
FIRST NUMBER =X................................1
HALF OF SECOND NUMBER =(X+1)/2...............2
TWICE THE THIRD =2(X+2)=2X+4
SEVEN LESS THAN THIS =2X+4-7=2X-3...........3
SUM OF THESE 3 IS
X+(X+1)/2 +2X-3=2101.....MULTIPLY BY 2 THROUGH OUT
2X+X+1+4X-6=4202
7X=4202-1+6=4207
X=4207/7=601
SO THE INTEGERS ARE X=601,X+1=602,X+2=603.
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