Question 31106
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:


{{{x + (1/2)y + (2z-7) = 2101}}}


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:


{{{x + (1/2)(x+1) + (2(x+2)-7) = 2101}}}


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


{{{x + (1/2)x + (1/2) + (2x+ 4 -7) = 2101}}}


{{{(7/2)x  - (5/2) = 2101}}}


{{{(7/2)x  = 4207/2}}}


{{{x = 601}}}


So, the integers are 601, 602, 603.