How do you solve #2y-9>5y+12#?

1 Answer
Dec 21, 2016

Answer:

#y < -7#

Explanation:

First, isolate the #y# term on one side of the inequality and the constants on the other side of the inequality while keeping the inequality balanced:

#2y - 9 - color(red)(2y - 12) > 5y + 12 - 9 - color(red)(2y - 12)#

#2y - 2y - 9 - 12 > 5y - 2y + 12 - 12#

#0 - 9 -12 > 5y - 2y + 0#

#- 9 -12 > 5y - 2y#

Now we can group the like terms:

#-21 > (5 - 2)y#

#-21 > 3y#

And lastly, we can solve for #y# while keeping the inequality balanced:

#(-21)/color(red)(3) > (3y)/(color(red)(3)#

#-7 > (color(red)(cancel(color(black)(3)))y)/color(red)(cancel(color(black)(3)))#

#-7 > y#

And to get this solution in terms of #y# we can reverse or "flip" the inequality:

#y < -7#