Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

lkomarci
2005-09-14 19:58:47

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

lkomarci
2005-09-14 19:55:27

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

wcy
2005-09-14 19:31:53

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

MathsIsFun
2005-09-14 19:07:18

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

notamathperson
2005-09-14 17:25:43

Yes, sorry I forgot to add the parentheses.

MathsIsFun
2005-09-14 16:11:04

For x^2 you can also use x - 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/(x+3x-10) ?


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

notamathperson
2005-09-14 11:52:06

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.

kylekatarn
2005-09-14 10:49:59

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 : )

kylekatarn
2005-09-14 10:32:37

x squared = x^2

notamathperson
2005-09-14 09:24:57

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

Board footer

Powered by FluxBB