SOLUTION: Solve {{{1/(x+2)- 2/x=3}}} I used LCD and got (1x/(x^2 +2x))- (2x +4)/(x^2 + 2x) =3 Then multiplied bith sides by x^2 -2x 3x^2 - 5x -4

Algebra ->  Expressions-with-variables -> SOLUTION: Solve {{{1/(x+2)- 2/x=3}}} I used LCD and got (1x/(x^2 +2x))- (2x +4)/(x^2 + 2x) =3 Then multiplied bith sides by x^2 -2x 3x^2 - 5x -4      Log On


   

Question 81347: Solve 1%2F%28x%2B2%29-+2%2Fx=3
I used LCD and got (1x/(x^2 +2x))- (2x +4)/(x^2 + 2x) =3
Then multiplied bith sides by x^2 -2x
3x^2 - 5x -4
Found 2 solutions by bucky, Edwin McCravy:
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
%281%2F%28x%2B2%29%29-+%282%2Fx%29=3
.
I used LCD and got:
.
%281x%2F%28x%5E2+%2B+2x%29%29-+%282x+%2B4%29%2F%28x%5E2+%2B+2x%29+=+3 <===
.
Then multiplied both sides by x%5E2+-+2x <=== should be + between terms, not -
.
You should now have:
.
x+-+2x+-+4+=+3%28x%5E2+%2B+2x%29
.
Multiply out the right side and you get:
.
x+-+2x+-+4+=+3x%5E2+%2B+6x
.
Combine the two terms on the left side that both contain x to get:
.
+-x+-+4+=+3x%5E2+%2B+6x
.
Add x + 4 to both sides to eliminate the terms on the left side. The result is:
.
0+=+3x%5E2+%2B+6x+%2B+x+%2B+4
.
Combine the two terms on the right side that contain x:
.
0+=+3x%5E2+%2B+7x+%2B+4
.
Transpose (switch sides):
.
3x%5E2+%2B+7x+%2B+4+=+0
.
This equation is now in standard quadratic form. It can be solved by graphing, by factoring,
or by using the quadratic formula. The more general approach is to use the quadratic
formula, but in this case factoring works. The left side of this equation factors to:
.
%283x+%2B+4%29%2A%28x+%2B+1%29=+0
.
Notice that this equation will be true if either of the factors on the left side equals zero
because zero times anything is zero. So, one at a time, set the two factors equal to
zero and solve for x.
.
3x+%2B+4+=+0
.
Subtract 4 from both sides:
.
3x+=+-4
.
Divide by 3:
.
x+=+-4%2F3+
.
That's one solution for x. Now set the other factor equal to zero:
.
x+%2B+1+=+0
.
Subtract 1 from both sides:
.
x+=+-1
.
That's the second solution for x.
.
In summary, the two solutions for x are -4%2F3+ and -1
.
You can check both of these by substituting them (one at a time) for x in the original
given equation and making sure that the equation still balances on both sides.
.
Hope this helps you to understand the problem and also to correct your minor mistake.
.
Cheers ...

Answer by Edwin McCravy(20063) About Me  (Show Source):
You can put this solution on YOUR website!

Solve 1%2F%28x%2B2%29-+2%2Fx=3
I used LCD and got (1x/(x^2 +2x))- (2x +4)/(x^2 + 2x) = 3
Then multiplied bith sides by x^2 -2x
3x^2 - 5x -4

You need to multiply on both sides to begin with, not
multiply on the left and fail to on the right.  You also
should not multiply the LCD out as you did.  Here is
step by step, what you should have:

         1       2
      ------- - --- = 3
       x + 2     x 

Put the 3 over 1 so everything will be a fraction:

         1       2     3
      ------- - --- = ---
       x + 2     x     1

Now get the LCD = x(x + 2) but DON'T multiply it out!
That was one thing you did wrong above.  Again, don't multiply the LCD out!

Instead put it over 1 like this, with a multiplication dot 
symbol · beside it:

 x(x + 2)    
---------- · 
     1       

Now place that whole thing before each terms on the left side AND the right side:

       x(x + 2)       1       x(x + 2)     2     x(x + 2)     3 
      ---------- · ------- - ---------- · --- = ---------- · ---
           1        x + 2         1        x         1        1 
 
Now you're ready to do some canceling. 

1. Cancel the black x + 2 into the red (x + 2)
2. Cancel the black x into the red x

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

Now notice that after canceling, you have only 1's in the denominators.
So you can erase all the 1's and like magic all the fractions are gone:

1. All that's left in the first term is the red x
2. All the's left in the second term is - (x+2)2
3. All that's left on the right side is x(x+2)3

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

It's customary to write numbers before parenthese, not after them,
so we write that as

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


I'll now dispense with the colors and make everything black:

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

Now we're finally ready to multiply things out, but not until now!

                                   x - 2x - 4 = 3x² + 6x

Can you finish solving that?  If not post again asking how.
I'll assume you can.   You get two solutions x = -4/3 and x = -1

[Neither of these answers cause any denominators to become zero
when substituted in the original equation, so they are both
solutions.  Had we gotten x = -2 we would have had to discard
it because it would have made the denominator x - 2 become 0.
Sometimes this is the case when we multiply through by the LCD,
so we should always look to make sure we aren't including an 
extraneous solution. I thought I'd better warn you about
extraneous solutions because you do get them sometimes, but
not in this example.]

Edwin