How do you solve for b? b+7>#3/2b#+17

2 Answers
MathFact-orials.blogspot.com
Dec 14, 2017

#b<-20#

Explanation:

#b+7>3/2b+17#

For ease, let's multiply through by 2 to get rid of the fraction:

#2b+14>3b+34#

Let's now group like terms:

#2bcolor(red)(-2b-34)+14>3bcolor(red)(-2b-34)+34#

#-20>b#

To show this graphically, draw an open circle around #-20# on a number line (to show that it isn't part of the solution), and then draw a ray heading off towards #- oo#.

Nlyyildirim
Dec 14, 2017

#b<-20#

Explanation:

#b+7>3/2#b#+17#

add #-17# both sides;
#b+7-17>3/2b##+17-17#
#b-10>3/2b# (add -b both sides)
#-b+b-10>3/2b-b#
#-10>1/2b#(multiply by 2 both sides)
#2*(-10)>2*(1/2b)#
#-20>b#
in other words;

#b<-20#

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