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**notamathperson****Member**- Registered: 2005-09-13
- Posts: 5

Hey I'm a college algebra student who really struggles with this subject. Any help I can get here will be greatly appreciated.

Here are a few problems I've had on my homework. I'd really like to know how to work the problems, so if someone could show me step by step, that'd be great.

1. y-2[3-(y+9)]=12-(y-7)

2. 6/x-2 - 1/x+5 = 2/x2+3x-10

3. x/x-1 +3 = 1/x-1

That's all for now. I'm not sure what the best way to indicate x squared (as in #2, I put x2), though. Thanks

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

x squared = x^2

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

y-2[3-(y+9)]=12-(y-7)

y-2[3-y-9)]=12-y+7

y-(2[3-y-9])=19-y

y-(2*3-2*y-2*9)=19-y

y-(6-2y-18)=19-y

y-6+2y+18=19-y

3y+12=19-y

3y+y=19-12

4y=7

y=7/4

...

now you try the other ones. they're a bit harder than the first one, but you can solve them!

Try! Learning maths is all about trying and trying again!

If you get stuck, post your work and we'll work on that : )

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**notamathperson****Member**- Registered: 2005-09-13
- Posts: 5

Thanks Kyle. Here's the work I have so far on the others

6/x-2 - 1/x+5 = 2/x^2+3x-10

6x+30/x^2+3x-10 - x-2/x^2+3x-10

5x+32/x^2+3x-10 = 2/x^2+3x-10

Should I cross multiply and get big numbers or is their a more efficient way?

The third problem:

x/x-1 +3 = 1/x-1

x/x-1 + 3/1 = 1/x-1

x/x-1 + 3x-3/x-1 = 1/x-1

4x-3/x-1 = 1/x-1

Same thing here.

*Last edited by notamathperson (2005-09-13 13:53:53)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

For x^2 you can also use x² - there are a few useful little symbols at the top under "Math Is Fun Forum", and I just drag my mouse across one then copy and paste it in.

I have been trying (unsuccessfully so far) to get a "LaTeX" to work on the forum. With this you could use special notation to show expressions nicely. I wil have another attempt at getting this to work soon.

I have some questions:

Prob 2: 2. 6/x-2 - 1/x+5 = 2/x2+3x-10

Should that be: 6/(x-2) - 1/(x+5) = 2/(x²+3x-10) ?

Prob 3: x/x-1 +3 = 1/x-1

Should that be: x/(x-1) +3 = 1/(x-1) ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**notamathperson****Member**- Registered: 2005-09-13
- Posts: 5

Yes, sorry I forgot to add the parentheses.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

OK, let me have a go at Prob 3: x/(x-1) +3 = 1/(x-1)

Multiply both sides by (x-1): x +3(x-1) = 1

Expand: x +3x-3 = 1

Combine: 4x -3=1

Add 3: 4x=4

Therefore: x=1

But I can't check my work, because 1/(x-1) = 1/0

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

x/(x-1) +3 = 1/(x-1)

subtract 1/(x-1) on both sides

(x-1)/(x-1) +3=0

1+3=0

contradiction

therefore no solution

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**lkomarci****Member**- Registered: 2005-08-24
- Posts: 23

i have to say i'm proud that even i can help out someone:

i got the solution to the 2 task...it actually couldn't be easier.

6/(x-2) - 1/(x+5) = 2/x^2+3x-10

6/(x-2) - 1/(x+5) = 2/(x-2)(x+5) /(x-2)(x+5)

6(x+5) - (x-2) = 2

6x+30-x+2=2

5x = -30 /:5

x = -6

see?? not so hard...the only thing that you HAVE to notice is x^2+3x-10, which can also be written in form of (x-2)(x+5)

when u multiply the first two variables of each bracket (x,x) you get the X^2, when you sum up the second two variables (-2, 5) you get the 3x, and when u multiply them (-2, 5) you get the -10.

that's the way this thing works. you practice that, you have to be experienced enough to spot this as soon as you look at the exercise.

hope i've been helpful

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**lkomarci****Member**- Registered: 2005-08-24
- Posts: 23

no no no no WCY.. not correct

watch:

x/(x-1) + 3 = 1/(x-1) / (x-1)

x + 3(x-1) = 1

x + 3x - 3 = 1

x+ 3x = 1 + 3

4x = 4 /:4

x = 1

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