Asked by mikey
please help!!!! how do I find three times the first of two consecutive integers is at least twelve more than four times the second. And to find the greatest possible integers.
Answers
Answered by
Reiny
Two consecutive integers: x, and x+1
3x > 4(x+1) + 12
3x > 4x + 16
-x > 16
x < -16
So choices would be -16, -15 ... -17, -16 .... -18, -17 ... etc
there are no positive integers that work.
3x > 4(x+1) + 12
3x > 4x + 16
-x > 16
x < -16
So choices would be -16, -15 ... -17, -16 .... -18, -17 ... etc
there are no positive integers that work.