Absolute Value equations: "if B greater than or equal to 0 then A=B or A=-B"

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dnymeyer

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Absolute Value equations: "if B greater than or equal to 0 then A=B or A=-B"

Hey guys, something I ran into that is throwing me through a loop a little. Looking for some input/explanation?

For Absolute Value equation solving

|A|=B

if B greater than or equal to 0 then A=B or A=-B
if B less than zero, then |A| = B has no solution.

This is kind of confusing to me. I understand that Absolute Value equations cannot be negative (as Absolute Value is defined as a "positive" number away from zero)So I understand for example that the following equation has no solution:

|2v-12| = -8
The answer is No Solution, as the Absolute Value cannot be solved to get a negative number.


My question is then for something like this example:

|6w+18| = 6

I understand that this can be solved for 6w+18=6 and the answer is w= -2.

However, the answer is actually w= -2, -4. Since you are also supposed to solve for 6w+18= -6.

I'm looking at the original question of |6w+18| = 6 and it seems to be that since the full equation 6w+18 is in the absolute value signs, the answer should ALWAYS be positive 6. and NEVER a negative number.

However, this is wrong. The answer is both w= -2 and w=-4.

w=-2 brings the equation answer to 6..... but w =-4 brings the equation answer to -6.

I am not understanding this (why the answer is allowed to be negative when the equation is in absolute value signage)

Sorry if this is confusing. But can anyone help me out?
 
If I'm given:

\(\displaystyle |6w+18|=6\)

I would first factor:

\(\displaystyle 6|w+3|=6\)

Divide through by 6:

\(\displaystyle |w+3|=1\)

\(\displaystyle |w-(-3)|=1\)

Now, when I look at this, I see "w is a number such that its distance on the number line from -3 is 1." There are two such numbers, on either side of -3, 1 unit away, namely -4 and -2.
 
MarkFL said:
If I'm given:

\(\displaystyle |6w+18|=6\)

I would first factor:

\(\displaystyle 6|w+3|=6\)

Divide through by 6:

\(\displaystyle |w+3|=1\)

\(\displaystyle |w-(-3)|=1\)

Now, when I look at this, I see "w is a number such that its distance on the number line from -3 is 1." There are two such numbers, on either side of -3, 1 unit away, namely -4 and -2.
Click to expand...

Okay. We haven't learned Factoring in my class yet, so this explanation is a little rough for me. I do understanding Factoring is finding a common multiplier and using it to simplify.

But given this same logic, when you look at
|2v-12|= -8
Why is the solution No Solution?

You could also factor this and follow your given steps. To get |v-6| = -4. Then you could solve this, presumably, for both a positive and negative number.

But the answer is No Solution.

So I guess I'm still trying to find an answer as to why absolute values could sometimes be solved for a negative number, and other times it is not.
 
dnymeyer said:

My question is then for something like this example:

|6w+18| = 6

I understand that this can be solved for 6w+18=6 and the answer is w= -2.

However, the answer is actually w= -2, -4. Since you are also supposed to solve for 6w+18= -6.

I'm looking at the original question of |6w+18| = 6 and it seems to be that since the full equation 6w+18 is in the absolute value signs, the answer should ALWAYS be positive 6. and NEVER a negative number.

However, this is wrong. The answer is both w= -2 and w=-4.

w=-2 brings the equation answer to 6..... but w =-4 brings the equation answer to -6.

I am not understanding this (why the answer is allowed to be negative when the equation is in absolute value signage)

Sorry if this is confusing. But can anyone help me out?
Click to expand...

Your equation, |6w+18| = 6, says that 6w + 18 is a number whose absolute value is 6.

There are two numbers whose absolute value is 6: 6 and -6.

So 6w + 18 can be either 6 or -6.

Solving 6w + 18 = 6, we get w = -2.

Solving 6w + 18 = -6, we get w = -4.

So those are the two solutions.

Checking each solution, for w = -2 the left side is |6(-2) + 18| = |6| = 6, so that is a solution; and for w = -4, the left side is |6(-4) + 18| = |-6| = 6, so that is also a solution.

The thing that has to be positive is the value of the absolute value -- not the quantity inside, or the variable, or anything else. I think you are allowing your attention to drift to the wrong things. If you just take it step by step as I did above, both in solving and in checking, and forget about general statements about something being positive, you can see it correctly.

Does that make it clearer?
 
dnymeyer said:
Okay. We haven't learned Factoring in my class yet, so this explanation is a little rough for me. I do understanding Factoring is finding a common multiplier and using it to simplify.

But given this same logic, when you look at
|2v-12|= -8
Why is the solution No Solution?

You could also factor this and follow your given steps. To get |v-6| = -4. Then you could solve this, presumably, for both a positive and negative number.

But the answer is No Solution.

So I guess I'm still trying to find an answer as to why absolute values could sometimes be solved for a negative number, and other times it is not.
Click to expand...

Don't worry about factoring; that's just an extra step you can take to avoid working with larger numbers, and can be mastered after you understand the basics. (And it can be dangerous with absolute values, if you don't think clearly.)

This equation says that 2v - 12 is a number whose absolute value is -8. But an absolute value can never be negative, so there is no solution.

Part of your problem is that you are using words incorrectly, which makes it hard to think clearly. You aren't "solving for a negative number"; you are solving an equation that says an absolute value is negative, which can't happen. In the other problem, you are solving and finding that x is a negative number, which is entirely unrelated. And in the process, you are using the fact that the argument of the absolute value (the number inside it) can be either negative or positive. That is yet another different fact.
 
dnymeyer said:

|A|=B

if B greater than or equal to 0 then A=B or A=-B

My question is then for something like this example:

|6w+18| = 6

I understand that this can be solved for 6w+18=6 and the answer is w= -2.
However, the answer is actually w= -2, -4. Since you are also supposed to solve for 6w+18= -6.
I am not understanding this (why the answer is allowed to be negative when the equation is in absolute value signage)
Sorry if this is confusing. But can anyone help me out?
Click to expand...

As you wrote above, if B greater than or equal to 0 then A=B or A=-B - we get 2 equations for the price of one.
Why? Because the equality works in _2_ cases, when A is B and when A is -B. Try it: |x| = 5. |5| = 5, |-5| = 5.
Note that the case A=-B is NOT |A| = -B. As you said, the result of |A| is not allowed to be negative. But A can be!

So, in case of |6w+18| = 6 we get 2 equations:
6w+18 = 6, w = -2
6w+18 = -6, w = -4
 
Ohhh okay, I see. My question seems silly now that I see where I was misunderstanding! :p:D
Thanks for the help!
 
dnymeyer said:
Hey guys, something I ran into that is throwing me through a loop a little. Looking for some input/explanation?

For Absolute Value equation solving

|A|=B

if B greater than or equal to 0 then A=B or A=-B
if B less than zero, then |A| = B has no solution.

This is kind of confusing to me. I understand that Absolute Value equations cannot be negative (as Absolute Value is defined as a "positive" number away from zero)So I understand for example that the following equation has no solution:

|2v-12| = -8
The answer is No Solution, as the Absolute Value cannot be solved to get a negative number.


My question is then for something like this example:

|6w+18| = 6

I understand that this can be solved for 6w+18=6 and the answer is w= -2.

However, the answer is actually w= -2, -4. Since you are also supposed to solve for 6w+18= -6.

I'm looking at the original question of |6w+18| = 6 and it seems to be that since the full equation 6w+18 is in the absolute value signs, the answer should ALWAYS be positive 6. and NEVER a negative number.

However, this is wrong. The answer is both w= -2 and w=-4.

w=-2 brings the equation answer to 6..... but w =-4 brings the equation answer to -6.

I am not understanding this (why the answer is allowed to be negative when the equation is in absolute value signage)

Sorry if this is confusing. But can anyone help me out?
Click to expand...
When you have |6w+18| = 6 you should initially ignore the 6w+18 and ask yourself " the absolute value of what equals 6" The answer to that is 6 and -6. You seem to be thinking that |6w+18| can be 6 or -6. You are correct that |6w+18| can not have solution if it equals -6. The thing is that it is NOT |6w+18| that we are letting be 6 or -6 but rather we are letting 6w+18 = 6 or -6. Let's see if I am correct. 6w+18 = 6 means that wherever you see 6w+18 you can replace it with 6. So is | 6 | = 6 a true statement. Of course it is. Now how about when 6w+18 = -6. This means you can place -6 wherever you see 6w+18. When I do that I get | -6 | = 6 which is also true. So you need to solve two equations to get w and these two equations are 6w+18 = 6 and 6w+18 = -6.

We are glad that you asked this question as you must be clear on this topic.
 
To generalize a bit on what I posted before, consider the following:

\(\displaystyle |ax+b|=c\)

Now, if \(\displaystyle c<0\), we know there are no solutions and we are done, so let's assume then that \(\displaystyle 0\le c\).

Factor:

\(\displaystyle |a|\left|x+\dfrac{b}{a}\right|=c\)

Divide through by \(\displaystyle |a|\)

\(\displaystyle \left|x+\dfrac{b}{a}\right|=\dfrac{c}{|a|}\)

\(\displaystyle \left|x-\left(-\dfrac{b}{a}\right)\right|=\dfrac{c}{|a|}\)

And so we see that \(\displaystyle x\) is some number whose distance from \(\displaystyle -\dfrac{b}{a}\) on the number line is \(\displaystyle \dfrac{c}{|a|}\), that is:

\(\displaystyle x=-\dfrac{b}{a}\pm\dfrac{c}{|a|}\)
 
You guys are great! So helpful and nice.

Just to summarize where I can see I was making my mistake:

The equation given is
[FONT=MathJax_Main]|[FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]w[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]18[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]6[/FONT][/FONT]​
Thus I am asking "absolute value of this equation is 6, so anything in the absolute value equation can equal a positive or negative number because in the end it is asking for only the absolute value ("positive")

So when we solve for both 6w + 18 = 6, AND 6w + 18 = -6, that is only a hypothetical equation to solve since |6w +18| = 6 would never equal -6. I am only solving it for -6 to find the negative answer within the absolute value equation.

That's probably confusing again. I tend to get wordsy but that's how it's making sense to me now. I was confusing that solving to get a -6 is actually what the answer could be, but in reality it's just a temporary equation to solve for the negative answer in the absolute value equation.
[FONT=MathJax_Main]
[/FONT]​
 
dnymeyer said:
You guys are great! So helpful and nice.

Just to summarize where I can see I was making my mistake:

The equation given is
[FONT=MathJax_Main]|[FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]w[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]18[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]6[/FONT][/FONT]​
Thus I am asking "absolute value of this equation is 6, so anything in the absolute value equation can equal a positive or negative number because in the end it is asking for only the absolute value ("positive")

So when we solve for both 6w + 18 = 6, AND 6w + 18 = -6, that is only a hypothetical equation to solve since |6w +18| = 6 would never equal -6. I am only solving it for -6 to find the negative answer within the absolute value equation.

That's probably confusing again. I tend to get wordsy but that's how it's making sense to me now. I was confusing that solving to get a -6 is actually what the answer could be, but in reality it's just a temporary equation to solve for the negative answer in the absolute value equation.
[FONT=MathJax_Main]
[/FONT]​
Click to expand...

I think you are looking at it correctly. We can have:

\(\displaystyle |\pm6|=6\)

So, when given:

\(\displaystyle |f(x)|=6\)

We can solve:

\(\displaystyle f(x)=\pm6\)

to get the solutions.
 
dnymeyer said:
You guys are great! So helpful and nice.

Just to summarize where I can see I was making my mistake:

The equation given is
[FONT=MathJax_Main]|[FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]w[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]18[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]6[/FONT][/FONT]​
Thus I am asking "absolute value of this equation is 6, so anything in the absolute value equation can equal a positive or negative number because in the end it is asking for only the absolute value ("positive")

So when we solve for both 6w + 18 = 6, AND 6w + 18 = -6, that is only a hypothetical equation to solve since |6w +18| = 6 would never equal -6. I am only solving it for -6 to find the negative answer within the absolute value equation.

That's probably confusing again. I tend to get wordsy but that's how it's making sense to me now. I was confusing that solving to get a -6 is actually what the answer could be, but in reality it's just a temporary equation to solve for the negative answer in the absolute value equation.
[FONT=MathJax_Main]
[/FONT]​
Click to expand...
You keep talking about |6w+18| equaling 6 or -6. If you keep thinking that then you may get confused and make mistakes. Again, it is 6w+18 that can be 6 or -6. You need to walk around today thinking about this until you understand it completely.
 
Jomo said:
You keep talking about |6w+18| equaling 6 or -6. If you keep thinking that then you may get confused and make mistakes. Again, it is 6w+18 that can be 6 or -6. You need to walk around today thinking about this until you understand it completely.
Click to expand...

Yes, that is what I meant by saying "solving for 6w + 18 = -6 is merely a temporary hypothetical equation used only for the means to solve for a negative input in the absolute value".

I just worded my thoughts in a much more confusing way. Haha :) Thank you for wording my thoughts in a more succinct and clear verbiage.
 
dnymeyer said:
Yes, that is what I meant by saying "solving for 6w + 18 = -6 is merely a temporary hypothetical equation used only for the means to solve for a negative input in the absolute value".

I just worded my thoughts in a much more confusing way. Haha :) Thank you for wording my thoughts in a more succinct and clear verbiage.
Click to expand...

There is nothing temporary or hypothetical about that equation. If your assignment had this equation instead of the original one, I doubt you would say "Wow, people, we have a hypothetical equation to solve!"

The original equation with || operation is very permanently equivalent to 2 equations without one.
 
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