How do you solve #-4( 3+ m ) + 8m < 4( m + 3)#?

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Answer 1 · 2016-10-22T15:10:41

#m# can have any value.

Treat the bigger/smaller than sign like an equal sign

#(-4 xx 3) +( -4 xx m) + 8m < 4xx m + 4 xx 3#

#-12 - 4m +8m < 4m + 12" "larr# move variables to one side

#-4m +8m -4m < 12+12#

#0 < 24#

There is no m-term left to solve for, but we have ended up with a true statement.

This is an indication that the equation will be true for any value of m.