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Nov 16th, 2003, 04:53 PM
#1
Thread Starter
Dazed Member
Combining like terms?
The problem is 5 - 1/3x6 + 3/x2-2x
I get all the way to 15x(x-2)(x-2) + x(x-2) + 9(x-2) / 3x(x-2)(x-2) but know i dont know what to do next. I know i can only cancel out common factors not terms. The numerator is still broken up into three terms so how do i combine like terms in the numerator?
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Nov 16th, 2003, 05:16 PM
#2
Retired VBF Adm1nistrator
Is this the equation you typed out originally ?
Code:
5 - 1
-----------
3 . ( 6 + 3 )
--------
x^2 - 2x
Microsoft MVP : Visual Developer - Visual Basic [2004-2005]
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Nov 16th, 2003, 05:22 PM
#3
Frenzied Member
Dilenger you've gone wrong somewhere. I suppose your equations equals zero and you want to find X
in the first equation X cannot equal zero to get zero, but in the 2nd one can, could you please re-phrase the question and use brackets.
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Nov 16th, 2003, 05:27 PM
#4
Thread Starter
Dazed Member
Sorry. It's 5 - 1/3x - 6 + 3/x2-2x.
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Nov 16th, 2003, 05:32 PM
#5
Frenzied Member
do you mean
5 - (1/3x) - 6 + [3/(x^2-2x)] = 0 ?
in that case X is 2.3134703 according to my calc
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Nov 16th, 2003, 05:38 PM
#6
Thread Starter
Dazed Member
5 - 1 / (3x-6) + 3 / (x2 - 2x)
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Nov 16th, 2003, 05:44 PM
#7
Frenzied Member
5 - 1 / (3x-6) + 3 / (x2 - 2x) =0
does not work for any value of X
think about it, as X tends to +0, the result tends to -infinity. As X tends to -0, the result tends towards +infinity. As X tends towards infinity (+ve or -ve) the result tends towards 5.
the closest it even gets to zero is:
-0.661705 when X is 1.029435
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Nov 16th, 2003, 05:50 PM
#8
Thread Starter
Dazed Member
Im not trying to find the value of x. It's just adding and subtracting rational expressions.
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Nov 16th, 2003, 05:52 PM
#9
Frenzied Member
to tidy it up? does it get tidier than that?
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Nov 16th, 2003, 05:58 PM
#10
Thread Starter
Dazed Member
Code:
5 -
1
-----------
(3x - 6)
+
3
-----------
x2 - 2x
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Nov 16th, 2003, 06:21 PM
#11
Frenzied Member
well I've got something, but I wouldn't be the least suprised if it's totally wrong:
5-[1/((3X-6)+3/(X^2-2X))]
As i think that if you have a fraction under a fraction you can bring the bottoms denominator to the very top:
= 5- (X^2-2X)/[(3X-3)-3]
= 5- (X(x-2)/[3+3(X-2)]
multiply everything be 3
= 15 - [X(X-2)]/[1+(X-2)]
then I get stuck
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Nov 16th, 2003, 07:20 PM
#12
Thread Starter
Dazed Member
5 - 1 / (3x - 6) + 3 / (x2 - 2x)
5(3x-6)(x2 - 2x) - (x2 - 2x) + 3(3x-6) / (3x-6)(x2 - 2x)
5(3x3 - 12x2 + 12x) - x(x-2) + 9(x-2) / 3x(x-2)(x-2)
15x(x-2)(x-2) - x(x-2) + 9(x-2) / 3x(x-2)(x-2)
Now here is where im not sure what to do.
Last edited by Dilenger4; Nov 16th, 2003 at 07:25 PM.
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Nov 17th, 2003, 11:27 AM
#13
Assumption???
5 - 1 / (3x - 6) + 3 / (x2 - 2x) = 0???
Start: 5 - 1 / (3x - 6) + 3 / (x2 - 2x)
Step 1: Multiply All terms by (3x-6)*(x2-2x)
::::: 5*(3x-6)*(x2-2x) - (1/(3x-6))*(3x-6)*(x2-2x) + 3*(1/(x2-2x))*(3x-6)*(x2-2x)
Step 2: Perform Multiplication on fractions:
:::::5*(3x-6)*(x2-2x) - (x2-2x) + 3*(3x-6)
Step 3: Result:
::::::15x3 - 61x2+71x-18 = 0
Now, let me check my math... oops!Bad Lou! -49 should be 71
So, Hows that!
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Nov 17th, 2003, 11:31 AM
#14
Now, if you mean to get a single fraction, then I think it is:
(15x3 - 61x2+71x-18)/(3x3-12x2+12x)
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Nov 17th, 2003, 01:58 PM
#15
Thread Starter
Dazed Member
Lou. I come up with your answer for the numerator if i take this approach.
5(3x-6)(x2 - 2x) - (x2 - 2x) + 3(3x-6) / (3x-6)(x2 - 2x)
5(3x3 - 12x2 + 12x) - (x2 - 2x) +(9x- 18) / 3x(x-2)(x-2)
15x3 - 60x2 + 60x - x2 + 2x + 9x -18
15x3 - 61x2 + 71x - 18
The answer is supposed to be 15x2 - 31x + 9 / 3x(x-2)
If i try to factor the resulting polynominal( 15x2 - 31x + 9) to (? ?)(? ?) it dosent work for obvious reasons.
So i think i was on the right path with the last step i was at. 15x(x-2)(x-2) - x(x-2) + 9(x-2) / 3x(x-2)(x-2). Very confusing.
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