is the inequality always sometimes or never true 3(2x+1)5x-(2-x)

Question

is the inequality always sometimes or never true 3(2x+1)>5x-(2-x)

zepdrix · Answer

Hey Austin! $\Large \color{royalblue}{\text{Welcome To OpenStudy! :)}}$

anonymous · Answer

never true i guess

zepdrix · Answer

$$\Large 3(2x+1)\gt5x-\color{#FF11AA}{(}2-x\color{#FF11AA}{)}$$Let's start by distributing the negative sign to each term in these pink brackets.
Which gives us,$$\Large 3(2x+1)\gt5x-2+x$$

zepdrix · Answer

Do you understand how to distribute the 3 on the left side of the equation?

anonymous · Answer

no how do i do that

zepdrix · Answer

$$\Large 3\color{#FF11AA}{(}2x+1\color{#FF11AA}{)}\gt5x-2+x$$To distribute, we'll multiply the 3 by each term in the pink brackets, as we do so, we can drop the brackets.$$\Large 6x+3\gt5x-2+x$$
Understand what I did there? $\large 3\cdot 2x=6x$ and $\large 3\cdot1=3$

zepdrix · Answer

From here, add the x-terms that are on the right side.

anonymous · Answer

soooo that side would say 6x-2?

zepdrix · Answer

yes good :)$$\Large 6x+3\gt 6x-2$$From here we can simplify further by subtracting 6x from each side.

anonymous · Answer

then i would get 1>-2 right?

zepdrix · Answer

Hmm how did your 3 change to a 1? :O
You should get 3 > -2

When is that inequality true?
Since the numbers are not `equivalent` it's either going to be `always` or `never`.

anonymous · Answer

i wrote the wrong number in thank you so my answer would be never right?

zepdrix · Answer

Hmm 3 is greater than -2 when?
I think it's `always` greater, isn't it? :o

anonymous · Answer

i didnt understand but i do now