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#1 2005-09-13 11:24:57

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

Homework problems

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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#2 2005-09-13 13:52:06

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

Re: Homework problems

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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#3 2005-09-13 18:11:04

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Homework problems

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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#4 2005-09-13 19:25:43

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

Re: Homework problems

Yes, sorry I forgot to add the parentheses.

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#5 2005-09-13 21:07:18

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Homework problems

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


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

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#6 2005-09-13 21:31:53

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

Re: Homework problems

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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#7 2005-09-13 21:55:27

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

Re: Homework problems

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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#8 2005-09-13 21:58:47

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

Re: Homework problems

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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