How do you solve #-2x + 3\geq 4x - 9#?

1 Answer
EZ as pi
Nov 1, 2016

#x <= 2#

Explanation:

#-2x + 3\geq 4x - 9#

Treat this is the same way as an equation unless you multiply or divide by a negative number in which case the sign in the middle changes around.
However, it is always possible to avoid having to do that if you keep the variable positive.

#cancel(-2x) + 3 cancel(+2x) \geq 4x - 9+2x" "larr# add #2x# to both sides

#3 >= 6x -9" "larr# add 9 to both sides

#3+9 >= 6x-9+9#

#12 >= 6x" "larr div 6#

#2 >= x#

This can also be written as #x <= 2#

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