How do you solve and write the following in interval notation: #8x -3x + 2< 2(x + 7)#?

1 Answer
Jun 2, 2016

Answer:

#(-oo,4)#

Explanation:

We have the inequality

#8x-3x+2 < 2(x+7)#

On the left hand side, the #8x# and #-3x# can be combined:

#overbrace(8x-3x)^(8x-3x=5x)+2 < 2(x+7)#

#5x+2 < 2(x+7)#

Next, on the right hand side, distribute the #2#:

#5x+2 < overbrace(2(x+7))^(2(x)+2(7))#

#5x+2 < 2x+14#

Subtract #2x# from both sides.

#overbrace(5x-2x)^(3x)+2 < overbrace(2x-2x)^0+14#

#3x+2 < 14#

Subtract #2# from both sides.

#3x+overbrace(2-2)^0 < overbrace(14-2)^12#

#3x < 12#

Divide both sides by #3#.

#(3x)/3 < 12/3#

#x < 4#

Now, we need to express #x<4# in interval notation. Since #x# has to be less than #4#, we know that #4# is the upper bound of the interval.

Since there is no lower bound, #x# can extend all the way down, from #4#, to #0#, into the negative numbers, working its way left on the number line all the way to negative infinity.

So, the interval is #(-oo,4)#. Note that #(# and #)# are used instead of #[# and #]# because #x<4#, not #x<=4#. Additionally, #(# and #)# are always used with #oo# because #x# can never equal #oo#.