SOLUTION: Hi! I'm really having difficulty solving these problems. :( How do you solve for x in these problems? Hope someone would help me. Thank you in advance! {{{(x+1)/(3x-6) = (5x)/

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi! I'm really having difficulty solving these problems. :( How do you solve for x in these problems? Hope someone would help me. Thank you in advance! {{{(x+1)/(3x-6) = (5x)/      Log On


   

Question 1094535: Hi! I'm really having difficulty solving these problems. :( How do you solve for x in these problems? Hope someone would help me. Thank you in advance!
%28x%2B1%29%2F%283x-6%29+=+%285x%29%2F6+%2B+1%2F%28x-2%29
%283x%29%2F%28x%2B1%29+-+5%2F%282x%29+=+3%2F%282x%29
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

%28x%2B1%29%2F%283x-6%29+=+%285x%29%2F6+%2B+1%2F%28x-2%29 

Factor the first denominator by factoring out 3

%28x%2B1%29%2F%283%28x-2%29%29+=+%285x%29%2F6+%2B+1%2F%28x-2%29

The LCD is 6(x-2) so we multiply every term
through by red%286%28x-2%29%2F1%29



We cancel whatever will cancel:



What we have left is:

2%28x%2B1%29=%28x-2%29%285x%29%2B6

2x%2B2=5x%28x-2%29%2B6
2x%2B2=5x%5E2-10x%2B6

Get 0 on the right:

-5x%5E2%2B8x-4=0

Multiply through by -1 so x2 term will be positive:

5x%5E2-8x%2B4=0

Factor:

%285x-2%29%28x-2%29=0

5x-2 = 0;   x-2 = 0
  5x = 2      x = 2
   x = 2%2F5

Two answers are 2%2F5 and 2.

However, the original problem has denominators that contain 
variables.  Therefore we must check for extraneous answers 
that are not solutions because they cause the denominator to 
equal 0.  No denominator can ever equal to 0.  

The denominators in the original equation are 3x-6 and x-2

Substituting 2%2F5 for x in denominator 3x-6 is

3%282%2F5%29-6%22%22=%22%226%2F5-6%22%22=%22%226%2F5-30%2F5%22%22=%22%22-24%2F5 which is not 0

Substituting 2%2F5 for x in denominator x-2 is

2%2F5-2%29%22%22=%22%222%2F5-10%2F5%22%22=%22%22-8%2F5 which is not 0 
either, so 2%2F5 is a solution..

Substituting 2 for x in denominator 3x-6 is

3%282%29-6%22%22=%22%226-6%22%22=%22%220 so we must

discard the answer x = 2, for it is extraneous and is not a
solution.   Therefore there is only one solution, x = 2%2F5

----------------------------------------

The second one:

%283x%29%2F%28x%2B1%29+-+5%2F%282x%29+=+3%2F%282x%29

Add %22%22%2B5%2F%282x%29 to both sides:

%283x%29%2F%28x%2B1%29+=+3%2F%282x%29+%2B+5%2F%282x%29

Since both terms on the right have the same denominator
we can just add the numerators and place it over their
common denominator:

%283x%29%2F%28x%2B1%29+=+8%2F%282x%29

Simplify the right side by dividing top and bottom by 4:

%283x%29%2F%28x%2B1%29+=+4%2Fx

We could get an LCD as we did in the other one but since 
there is only a fraction on each side, we can just 
cross-multiply:

3x%28x%29+=+4%28x%2B1%29

3x%5E2+=+4x%2B4

Get 0 on the right:

3x%5E2-4x-4=0

%283x%2B2%29%28x-2%29=0

3x+2 = 0;    x-2 = 0
  3x = -2;     x = 2
   x = -2%2F3

Two answers are -2%2F3 and 2.

However, the original problem also has denominators that contain 
variables.  Therefore we must check for extraneous answers 
that are not solutions because they cause the denominator to 
equal 0.  No denominator can ever equal to 0.  

The denominators in the original equation are x-1 and 2x.

Substituting -2%2F3 for x in denominator x-1 is

-2%2F3-1%22%22=%22%22-2%2F3-3%2F3%22%22=%22%22-5%2F3
which is not 0

Substituting -2%2F3 for x in denominator 2x is

2%28-2%2F3%29%22%22=%22%22-4%2F3 which is not 0 either, so 
-2%2F3 is a solution.

Substituting 2 for x in denominator x-1 is

2-1%22%22=%22%221which is not 0

Substituting 2 for x in denominator 2x is

2%282%29%22%22=%22%224 which is not 0 either, so 
2 is also a solution.

So there are two solutions  -2%2F3 and 2. 

But as you see we must make sure that any answer we get
really is a solution by substituting into each denominator
to show that no answer causes any denominator in the original
equation to be 0.

Edwin