How do you solve #\frac{- 4p - 1}{2} \geq 9#?

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Answer 1 · 2018-07-18T15:59:33

#p\le-19/4#

#\frac{-4p-1}{2}\ge 9#

#\frac{-4p-1}{2}\times 2\ge 9\times 2#

#-4p-1\ge 18#

#-4p-1+1\ge 18+1#

#-4p\ge19#

#-{4p}/(-4)\le19/(-4)#

#p\le-19/4#

Answer 2 · 2018-07-18T22:55:54

#p<= -19/4#

Let's multiply both sides by #2# to get rid of the fraction. We now have

#-4p-1>=18#

Adding #1# to both sides, we get

#-4p>=19#

Next, we divide both sides by #-4#. Since we are applying a negative to the inequality, we flip the sign. We now have

#p<= -19/4#

Hope this helps!