How do you solve #12> -3q# and #-2q> -12#?

1 Answer
Nov 3, 2017

Answer:

Very carefully! #-4< q <6#

Explanation:

You need to be careful when dealing with inequalities and negative coefficients, as multiplying or dividing by a value which might be negative can invalidate the inequality.
The safest method is to add the variable terms to both sides to make their coefficients positive first:
#12 > -3q rArr 12+3q > 0 rArr 3q > -12 rArr q > -4#
and:
#-2q > -12 rArr 0 > 2q-12 rArr 12 > 2q rArr 6 > q#
We can combine these two in a single interval:
#-4 < q < 6#