How do you solve #\frac { x - 1} { 2} \geq \frac { x } { 3} + 2#?

1 Answer
Azimet
Feb 1, 2017

#x >= 15#

Explanation:

To get rid of the denominators, you could multiply by #2# or #3#, but this would only take care of one problem. Because of this, its typically best to multiply both sides by (2 and 3's) least common factor, which is #6#:

#(6(x-1))/2 >= (6x)/3 + 12#

Since #6 > 0#, the inequality remains the same. If we multiplied (or divided) by a negative number, it would be reversed. Then, we can do some simplifications on the fractions, because

#6/2 = 3# and #6/3 = 2#.

#3(x-1) >= 2x + 12#

#3x - 3 >= 2x + 12#

#3x - 2x cancel(-3) cancel(+3) >= cancel(2x) cancel(-2x) + 12 + 3#

#x >= 15#

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