December 15, 2019, 08:47:04 pm

### AuthorTopic: Can anyone help me?  (Read 1072 times) Tweet Share

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#### elaine

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##### Can anyone help me?
« on: November 21, 2007, 09:21:32 pm »
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man I can't even do the revision questions

Solve for x

x/(a-b)   +   2x/(a+b)   =    1/ (a^2-b^2)

and another one

Solve for x

1/(x+a) + 1/(x+2a)  =  2/(x+3a)

can anyone show me the steps? sorry it's in an annoying calculator form

thanks
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#### asa.hoshi

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##### Can anyone help me?
« Reply #1 on: November 21, 2007, 09:55:34 pm »
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x/(a-b) + 2x/(a+b) = 1/ (a^2-b^2)
-->find common denominator
[x(a+b) + 2x(a-b)]/[(a-b)(a+b)] = 1/ (a^2-b^2)
note:(a-b)(a+b) - difference of 2 squares rule = a^2-b^2
[ax+bx+2xa-2bx]/a^2-b^2 = 1/ (a^2-b^2)
[3xa - bx]/(a^2-b^2) = 1/ (a^2-b^2)
both have same denominator
3xa - bx = 1
take x out as the factor
x(3a - b) = 1
therefore
x = 1/(3a - b)
i think thats right...
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#### asa.hoshi

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##### Can anyone help me?
« Reply #2 on: November 21, 2007, 10:04:22 pm »
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similar method as above but you are not dealing with difference of 2 squares.
1/(x+a) + 1/(x+2a) = 2/(x+3a)
[1(x+2a)+1(x+a)]/[(x+a)(x+2a)]  = 2/(x+3a)
(x+2a+x+a)/[(x+a)(x+2a)]= 2/(x+3a)
(2x+3a)/[(x+a)(x+2a)]= 2/(x+3a)
cross-multiplication
(2x+3a)(x+3a) = 2(x+a)(x+2a)
expand
2x^2+6ax+3ax+9a^2 = 2[x^2+2ax+ax+2a^2]
2x^2+9ax+9a^2 = 2x^2 + 6ax + 4a^2
simplify
2x^2+9ax+9a^2 - (2x^2 + 6ax + 4a^2) = 0
3ax + 5a^2 = 0
3ax = -5a^2
x = -5a/3

i think that is right.
I KNOW WHAT YOU DID LAST SUMMER!!

#### elaine

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##### Can anyone help me?
« Reply #3 on: November 21, 2007, 10:07:33 pm »
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hey thanks so much for your help
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#### asa.hoshi

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##### Can anyone help me?
« Reply #4 on: November 21, 2007, 10:08:45 pm »
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np.its a good way to get my post count up
I KNOW WHAT YOU DID LAST SUMMER!!

#### elaine

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##### Can anyone help me?
« Reply #5 on: November 21, 2007, 10:10:43 pm »
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lol here's another one for you- how does cross producting work?
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#### asa.hoshi

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##### Can anyone help me?
« Reply #6 on: November 21, 2007, 10:24:46 pm »
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x/y = w/z
u use cross-multiplication to eliminate the denominators.
so you take the denominator of one side to the other side of the equation and vice versa
x/y = w/z ==> xz = wy

but mathematically this is what really happens:
x/y = w/z
eliminate the LHS denominator. so multiple the LHS by "y". Whatever you do to the LHS you do the RHS so the equation does not change in value.
x/y multiples by y = w/z  multiples by y
y cancels out in the LHS
x = wy/z
do the same with the RHS
x multiples by z = wy/z multiples by z
z cancels out in the RHS
therefore xz = wy
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#### enwiabe

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##### Can anyone help me?
« Reply #7 on: November 21, 2007, 10:57:59 pm »
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That's not a cross product. Ew, don't get the real vector cross product confused with equation solving using cross-multiplication. That's got to be the equivalent of mathematical blasphemy

#### asa.hoshi

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##### Can anyone help me?
« Reply #8 on: November 21, 2007, 11:07:29 pm »
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sorry...
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#### elaine

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##### Can anyone help me?
« Reply #9 on: November 22, 2007, 12:41:24 pm »
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Quote from: "enwiabe"
That's not a cross product. Ew, don't get the real vector cross product confused with equation solving using cross-multiplication. That's got to be the equivalent of mathematical blasphemy

um i have no idea what that means lol.
but i get how the cross product/cross multiplication thingo works. thanks
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#### Collin Li

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##### Re: Can anyone help me?
« Reply #10 on: December 01, 2007, 06:26:55 pm »
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Okay. It's called cross-multiplication, because you have something like this:
$\frac{a}{b} = \frac{c}{d}$

Now, you multiply both sides by b, and you multiply both sides by d. Effectively, you bring the b to the RHS, and the d to the LHS.
$\implies a\times d = b\times c$

That is why it's called cross multiplying, because you're taking across the denominators to the other side (by multiplication).

The "cross product" is a completely different operation. Don't worry about it.
« Last Edit: January 27, 2008, 10:45:10 am by coblin »