Question 242519
the largest 3 consecutive integers for which the first is increased by twice the second exceeds the third by less than 25. (have to solve algrebraically)

I do not know how to put "exceeds" in an equation.


Let the 1st of the 3 consecutive integer be F


Then the 2nd and 3rd integers are F + 1, and F + 2, respectively


Increasing the 1st by twice the second gives us F + 2(F + 1). Since this result exceeds the third by less than 25, it means that F + 2(F + 1) is greater than the 3rd integer, and when the 3rd integer, or F + 2 is subtracted from F + 2(F + 1), we should get a number that is less than 25. Therefore, putting this in mathematical terminology, we get:

F + 2(F + 1) - (F + 2) < 25


F + 2F + 2 - F - 2 < 25


2F < 25


Therefore, F, or the 1st number < {{{25/2}}}, or < 12.5


Since we're talking about INTEGERS, then the integer that is less than 12.5 is 12. Therefore, with F, or the 1st integer being 12, the 3 INTEGERS are {{{highlight_green(12_13_and_14)}}}.