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

doubleu

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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.


 

galactus

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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.
 

Subhotosh Khan

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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.


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
 

doubleu

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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".



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)
 

galactus

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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".

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)
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.
 

doubleu

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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.
 

galactus

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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.
 

doubleu

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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.
 

Subhotosh Khan

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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'
 

doubleu

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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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