I need help solving this equation 2/(x+1)=x/(3-2x)

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veronicadeno

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The question says to use my factoring skills to solve this equation.

2/(x+1)=x/(3-2x)

(The / is for division, I don't want to confuse anybody.)
 
Start with Domain issues

x is NOT -1 and x is NOT 3/2.

Why did I do that and now what?

{repaired}
 
Well then, you cannot adequately solve this equation.

Generally, when hoping to get things out of a denominator, it requires multiplications.

3 = x/4

Multiply by 4, right?

It's not so obvious when the object in the denominator is not a constant.

3 = 4/x

Should you multiply by x?
 
veronicadeno said:
The [instructions say] to use my factoring skills to solve this equation.

2/(x + 1) = x/(3 - 2x)
Click to expand...

I'm thinking that the factoring skills do not apply to the given equation, but rather to the quadratic equation into which form the given equation can be put.

Multiplying both sides of the given equation by the LCD clears the ratios, leaving you with one polynomial equal to another. Next, subtracting the polynomial on the right from both sides yields a quadratic polynomial equal to zero.

Then, use your factoring skills to solve for x, and check your answers to be sure that they satisfy the original equation.

(Multiplying both sides of the given equation by the LCD has the effect of "cross-multiplication"; perhaps, you're familiar with this process.)

 
Re:

mmm4444bot said:
veronicadeno said:
The [instructions say] to use my factoring skills to solve this equation.

2/(x + 1) = x/(3 - 2x)
Click to expand...
Click to expand...
Or, use cross-multiplication; RULE: if a/b = (c/d), then ad = bc
So:
2/(x + 1) = x/(3 - 2x)

x(x + 1) = 2(3 - 2x) ; .......happy "factoring" :wink:
 
Re: Re:

Or, use cross-multiplication; RULE: if a/b = (c/d), then ad = bc
Click to expand...

GAAAaaaa!!!! ONLY if a/b exists and c/d exists. Or if b is NOT zero and d is NOT zero.

One must be very careful when making the jump from constants to non-constants.
 
RULE: if a/b = (c/d) "where b<>0 and d<>0", then ad = bc

Before Subhotosh gets a hold of this....I'm heading for the corner :(

Seriously: of course, denominators cannot equal 0; but where is the line drawn:

(x-1) / (x-4) = (x-2) / (x-3) where x <> 3 or 4
Re-arrange:
(x-1) / (x-2) = (x-4) / (x-3) where x <> 3 or 2 ; so x can now equal 4
Re-arrange:
1 / [(x-2)(x-4)] = 1 / [(x-1)(x-3)] where x <> 1,2,3,4
Keep on...
 
Thank you for your help! I asked my teacher today and she told me to cross multiply then set it equal to zero and factor.
So,
2/(x+1)=x/(3-2x)
x^2+x=6-4x
x^2+5x-6=0
x^2-x+6x-6=0
x(x+1)+6(x-1)=0
(x-1)(x+6)=0
x-1=0 x+6=0
x=1 x=-6
 
No mention of Domain Issues?
No demand to CHECK the result?

What has happened to the thorough math teacher? :cry:
 
veronicadeno said:
The question says to use my factoring skills to solve this equation.
2/(x+1)=x/(3-2x)
Click to expand...
Funny/strange how this format created all kinds of "philosophy";
but if presented this way:
solve x^2 + 5x - 6 = 0 by factoring
then all's well that ends well....
 
...except that it's a different problem with a different Domain - but you new I was going to say that. :wink:

I thought it ironic that you used the in-line division symbol in your statement.
 
tkhunny said:
I thought it ironic that you used the in-line division symbol in your statement.
Click to expand...
Whoops....funny/(strange) where s,t,r,a,n,g,e <> 0 8-)
 
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