How to solve |2x-8| < 10 and another

doubleu

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Aug 19, 2007
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Hi I'm stuck on the following two problems and would like to ask for help on how to solve them.
Any help will be much appreciated.


4lrbrro.jpg
 
For the first one.

Divide through by 2 and get \(\displaystyle \L\\|x-4|<5\)

Rewrite as \(\displaystyle \L\\-5<x-4<5\)

Now, solve for the interval by adding 4.


For the second one:

\(\displaystyle \L\\\frac{1}{y+1}+a=\frac{3}{y+1}\)

multiply through by the LCD, y+1, to eliminate the fractions. Then solve for y.
 
doubleu said:
Hi I'm stuck on the following two problems and would like to ask for help on how to solve them.
Any help will be much appreciated.


4lrbrro.jpg

Hint:

#1

(2x-8) < 10....................and.................. -(2x - 8) < 10

Now solve for 'x' from both the conditions and combine your answer.

#2

multiply both sides by (y+1) and solve for 'y'

state that y cannot equal to -1
 
galactus said:
For the first one.

Divide through by 2 and get \(\displaystyle \L\\|x-4|<5\)

Rewrite as \(\displaystyle \L\\-5<x-4<5\)

Now, solve for the interval by adding 4.
I'm confused on the rewrite part how did you get the -5 on the left side?

This is what I got from "divide by 2" then "add 4".

5x812s3.jpg


For the second one...
I'm not sure how to multiply by (y+1)
I know when I write it, it should be like this...but where do I go from there?
(very confused)
4u94cg6.jpg
 
I'm confused on the rewrite part how did you get the -5 on the left side?

That's a property of absolute values. \(\displaystyle \L\\|a|<b\) is equivalent to \(\displaystyle \L\\-b<a<b\)

This is what I got from "divide by 2" then "add 4".

5x812s3.jpg

Just add 4 throughout. \(\displaystyle \L\\-5<x-4<5\)

\(\displaystyle \L\\-1<x<9\)

For the second one...
I'm not sure how to multiply by (y+1)
I know when I write it, it should be like this...but where do I go from there?
(very confused)
4u94cg6.jpg

You forgot to multiply a by y+1 also

\(\displaystyle \L\\(y+1)\cdot\frac{1}{y+1}+a(y+1)=(y+1)\cdot\frac{3}{y+1}\)

Cancel accordingly and solve for y.
 
After cancelling y+1 I get 1+a=3. Subtract 1 from both sides and get a=2. Subtract a from both sides and have 2-a.
Not sure if I did it correctly.
 
No.

You cancelled too many y+1.

You should have \(\displaystyle \L\\1+a(y+1)=3\)

Don't cancel the y+1 if there's no division.

Okily-dokily.
 
galactus said:
No.

You cancelled too many y+1.

You should have \(\displaystyle \L\\1+a(y+1)=3\)

Don't cancel the y+1 if there's no division.

Okily-dokily.
Sorry but I'm even more confused now. :(
So that would be the answer or I have to foil it?
The solution in this packet says the answer to it is 2-a/a
I'm not sure how to end up with that answer from what I've tried with the help.
 
You really need to review pre-algebra:

1 + a (y+1) = 3

1 + ay + a = 3 ..............Distribute (ie, multiply out)

ay = 3 - 1 - a..............Isolate 'y'

ay = 2 - a..............Simplify


y = (2-a)/a..............Isolate 'y'
 
Yes I haven't taken math for so long so completing this review packet has been a challenge. I appreciate all the help thank you very much.
 
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