Absolute Value Equation--What Did I do Wrong?

[Hope101914]

Hello! I have this absolute value equation that I completed and one of my answers was marked as wrong. I have re-tried the problem several times and got the same answer. I even tried plugging it back into the equation and it came out right. I can't figure out what I did wrong. Any help would be greatly appreciated. Here's the problem and my work.

|x+4| = 5x-2

x+4 = 5x-2
6 = 4x
3/2 = x

x+4 = -5x+2
6x = -2
x = -1/3

My teacher said x=3/2 was correct but x=-1/3 was not. Did I do something wrong? Or did he possibly make a mistake?
Thanks!

 

[pka]

Hello! I have this absolute value equation that I completed and one of my answers was marked as wrong. I have re-tried the problem several times and got the same answer. I even tried plugging it back into the equation and it came out right. I can't figure out what I did wrong. Any help would be greatly appreciated. Here's the problem and my work.
|x+4| = 5x-2
x+4 = 5x-2
6 = 4x
3/2 = x

x+4 = -5x+2
6x = -2
x = -1/3

My teacher said x=3/2 was correct but x=-1/3 was not. Did I do something wrong? Or did he possibly make a mistake?

No, he is correct: \(\displaystyle \frac{-3}{2}\) gives a positive on the right and a negative on the left. Can't happen!

You need to solve \(\displaystyle x^2+8x+16=25x^2-20x+4\)

 

[Ishuda]

Hello! I have this absolute value equation that I completed and one of my answers was marked as wrong. I have re-tried the problem several times and got the same answer. I even tried plugging it back into the equation and it came out right. I can't figure out what I did wrong. Any help would be greatly appreciated. Here's the problem and my work.

|x+4| = 5x-2

x+4 = 5x-2
6 = 4x
3/2 = x

x+4 = -5x+2
6x = -2
x = -1/3

My teacher said x=3/2 was correct but x=-1/3 was not. Did I do something wrong? Or did he possibly make a mistake?
Thanks!

When you write
x+4 = 5x -2
what you are implying is that x+4 is positive and also 5x-2 is positive. Putting these together we
x > -4 and x > 2/5
This is true if x is greater than 2/5. Now you solve the equation for x and find you get x = 3/2. That is greater than 2/5 so everything is o.k. and x = 3/2 is a good answer.

Now look at the other side. When you write
x+4 = -5x + 2
you are implying -(x+4) and 5x -2 are positive [or, equivalently, x+4 and -5x+2 are negative] which says
x < -4 and x > 2/5
Well x can't be both less than -4 and greater than 2/5 so there is no solution for |x+4| = -(x+4)

Another way to check is to put, first, 3/2 in the original equation:
|3/2+4| = 5.5
and
5 * 3/2 - 2 = 5.5
It checks out.

Next, try -1/3
|-1/3 + 4| = \(\displaystyle \frac{11}{3}\)
and
5 * (-1/3) - 2 = -\(\displaystyle \frac{22}{3}\) EDIT: Someone can't add. I won't mention any names but the initials are Ishuda
It doesn't check out.

 

[Hope101914]

When you write
x+4 = 5x -2
what you are implying is that x+4 is positive and also 5x-2 is positive. Putting these together we
x > -4 and x > 2/5
This is true if x is greater than 2/5. Now you solve the equation for x and find you get x = 3/2. That is greater than 2/5 so everything is o.k. and x = 3/2 is a good answer.

Now look at the other side. When you write
x+4 = -5x + 2
you are implying -(x+4) and 5x -2 are positive [or, equivalently, x+4 and -5x+2 are negative] which says
x < -4 and x > 2/5
Well x can't be both less than -4 and greater than 2/5 so there is no solution for |x+4| = -(x+4)

Another way to check is to put, first, 3/2 in the original equation:
|3/2+4| = 5.5
and
5 * 3/2 - 2 = 5.5
It checks out.

Next, try -1/3
|-1/3 + 4| = \(\displaystyle \frac{11}{3}\)
and
5 * (-1/3) - 2 = -\(\displaystyle \frac{22}{3}\)
It doesn't check out.

So you are saying that -1/3 is an extraneous solution and 3/2 is the only solution?

Also, when I plug it in the the previous equation I get what Denis previously said:

-1/3 + 4 = 11/3
5(-1/3) - 2 = -11/3

Not -\(\displaystyle \frac{22}{3}\). Why am I getting a different answer than you are? And if it is what I am getting, does that mean it is extraneous because one is positive and the other is negative?

Thanks!

 

[Ishuda]

So you are saying that -1/3 is an extraneous solution and 3/2 is the only solution?

Yes, it is an extraneous solution such as one can get if they are not careful solving equations: For example, suppose
x = y, then
1 = y/x, then let y = 0 and we get
1 = 0 or adding 1 to each side
2 = 1 = 0 or ...
n = n-1 = n-2 = n-3 = ... = 2 = 1 = 0
for all n and thus all integers are equal.

Also, when I plug it in the the previous equation I get what Denis previously said:



Not -\(\displaystyle \frac{22}{3}\). Why am I getting a different answer than you are? And if it is what I am getting, does that mean it is extraneous because one is positive and the other is negative?

Thanks!

Well, although I do hate to admit it, sometimes I get sloppy or maybe its just a lapse a memory and I forget how to add or ... Take your pick

 

[Hope101914]

Yes, it is an extraneous solution such as one can get if they are not careful solving equations: For example, suppose
x = y, then
1 = y/x, then let y = 0 and we get
1 = 0 or adding 1 to each side
2 = 1 = 0 or ...
n = n-1 = n-2 = n-3 = ... = 2 = 1 = 0
for all n and thus all integers are equal.



Well, although I do hate to admit it, sometimes I get sloppy or maybe its just a lapse a memory and I forget how to add or ... Take your pick

Ok, thanks, and don't worry about it, we all make mistakes sometimes. As I guess I did when I didn't mark it as extraneous cause it came out right except the sign. Thanks so much for your help!