Question

2. Solve and graph the inequality |4 – v| < 5.
a. Write the inequality as two inequalities without absolute value.
b. Solve the inequality and write the solution set.
c. Graph the solution on a number line.

Please help explain the process in a very simple way. I don't quite get it and I have read over this lesson many times.

Answer 1

Hey there! Welcome to OpenStudy! If you need any help with navigating the site or it's rules just shoot me a message.

Now on to your problem, for part (a) do you know what it is asking?

Answer 2

Not really /:

Answer 3

\(\bf |4 - v| < 5\implies
\begin{cases}
+(4 - v) < 5\\ \quad \\
\bf -(4 - v) < 5
\end{cases}\)

Answer 4

Okay, lets say you have this inequality,

\(|x|<4\)

An absolute value is a magnitude, how far is that number from zero.
So this means that, that positive number is x from zero and that negative is also x from zero. Make sense?

So we can then re-write it as,

\(~~~~~~~~x<4\\
-(x)<4\)

Now that jdoe has done it, do you see how he did so?

Answer 5

So basically, part 'a' is asking to write it how it would look if the solution were to be on the positive or negative (two inequalities)? <if that makes sense

Answer 6

Yes, for an absolute value you will always have two equations/inequalities. The positive version of what is in the abs. value as well as a negative version.

For part (b) what you need to do is solve for x in each of the inequalities. Do you think you could do that for me?

Answer 7

I'm a bit stumped at this part. Wouldn't the solution (in my problem) just be " v < 5 "
or in your given example problem, wouldn't it just stay x < 4?

Answer 8

Lets look at what jdoe did, which is completely correct.

He has,

\(\bf |4 - v| < 5\implies
\begin{cases}
+(4 - v) < 5\\ \quad \\
\bf -(4 - v) < 5
\end{cases}\)

So you would need to solve the two inequalities to the right.

\(~~~~~~~~4-v<5\\
-(4-v)<5\)

Now keep in mind for the second one that you need to distribute that negative.

Answer 9

I am still really confused. Am I trying to find the value of '4 - v' or just 'v'?

Answer 10

You are trying to solve for v by itself.

\(v<\cdot\cdot\cdot\)

Answer 11

I apologize. I don't know why my brain just won't process this.

Answer 12

Okay, so it would be 'v < 5' ?

Answer 13

No need, this is clearly new material and new material is always tough. I understand :)

Answer 14

Lets look at the first example shall we?

\(4-v<5\)

First we would want to get the variable by itself, we would do so by subtracting 4 from both sides.

\(-v<5-4\\
-v<1\)

Now, we want it to be positive v, so we would divide both sides by negative one.

\(\displaystyle \frac{-v}{-1}<\frac{1}{-1}\\
~\\
v>-1\)

And you have to remember to flip the sign whenever you multiply or divide by a negative number.

Does this make sense to you?

Answer 15

Ohhh! That makes sense. So, that limits the solution set down to anything greater than -1 and less than 5 on a number line?

Answer 16

(Btw sorry it took so long to reply, I was going to draw a number line as an example of how I would graph it but it was difficult)

Answer 17

Um, not quite. Try and solve the second inequality.

Answer 18

Oh. Now I'm lost again. How exactly do I solve 'v'?

Answer 19

\(-(4-v)<5\)

Use the same process that I did before, but be sure to distribute that negative across that sum first :)

Answer 20

I'm confused on the process that you used before. It says '4 - v < 5' so how did you know to subtract 4 from 5?

Answer 21

Because you need to get terms with a variable on one side and constants on the other. So you can do that by subtracting a term from both sides, or adding. Whatever you do to one side you have to do to the other usually.

Answer 22

Okay so with the negative inequality would I add 4 to 5?

Answer 23

Yes you would!

Answer 24

good job helpin @AshB austin :D

Answer 25

Awesome! So:
v < 5+4
v < 9
Then,
\[\frac{ x }{ 9 } < \frac{ 1 }{ 9 }\] or would it be \[\frac{ x }{ 9 } > \frac{ 1 }{ 9 }\]
?

Answer 26

You just have,

\(v>-1\)
\(v<9\)

Those are what you have, now do you think you could put those in a solution set?

Answer 27

Ohh okay.
and a solution set is like {0, 1, 2, etc.} correct? << An example, not my answer.

Answer 28

I think it is more like,

\((-\infty,-1]\cup[9,\infty)\)

I am not quite sure how it would work, I am not so familiar with these.

Answer 29

So the solution set would contain all numbers and fractions greater than (but not equal to -1) and less than (but not equal to) 9?

Answer 30

Indeed, now do you think you could put this on a number line?