How do you solve #-9< \frac{ x }{ 5} - 2#?

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Answer 1 · 2018-04-07T17:25:06

#x > - 35#

Start out by adding 2 to both sides to add like terms:

#-9 + 2 < x/5 cancel(-2 + 2)#

# - 7 < x/5#

Now multiply #5# to both sides to remove the division factor:

#-7*5 < x/cancel5 * cancel5#

#-35 < x#

You see how this is saying "#x# is greater than #-35#?" I'm just going to rearrange the equation so that this fact is more obvious:

#-35 < x#

#x > - 35#

This did not change the value of the expression since it still means "#x# is greater than #-35#.