SOLUTION: Hi. I'm trying to solve 2xy+5x-2y-5 over 3xy+4x-3y-4. Thank you!
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Question 274297: Hi. I'm trying to solve 2xy+5x-2y-5 over 3xy+4x-3y-4. Thank you!
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The key to this problem is factoring
Let's factor
Start with the given expression.
Group the terms.
Factor out the GCF 'x' from the first group.
Factor out the GCF -1 from the second group.
Factor out the GCF from the entire expression.
So factors to
--------------------------------
Now let's factor
Start with the given expression.
Group the terms.
Factor out the GCF 'x' from the first group.
Factor out the GCF -1 from the second group.
Factor out the GCF from the entire expression.
So factors to
------------------------------------------------
Now that we have the factorizations, we can now simplify:
Start with the given expression.
Factor the numerator (see the first factorization above)
Factor the denominator (see the second factorization above)
Highlight the common terms.
Cancel out the common terms.
Simplify.
So simplifies to
In other words,
Note: One thing to point out is that the value of 'x' does not affect the final outcome of the expression since the 'x' terms cancel out. This isn't so obvious when you look at the original expression, but it becomes clear when we reach the last step.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Then you have an extraordinarily difficult task ahead of you, one for which I cannot provide any assistance nor can anyone else. The problem is you cannot "solve"
You cannot solve anything that does not have an equals sign or some sort of relational operator.
It is a rather complex rational expression that can be simplified. If that is what you meant, then follow along:
Start with the numerator polynomial. Factor an 
out of the first two terms and factor a -1 out of the last two terms:
\ -\ 1\left(2y\ +\ 5\right)}{3xy\ +\ 4x\ -\ 3y\ -\ 4})
Next, factor 
from the two terms that remain:
\left(2y\ +\ 5\right)}{3xy\ +\ 4x\ -\ 3y\ -\ 4})
Next work on the denominator. Use a similar strategy, factoring out of the first two terms and -1 out of the last two terms:
\left(2y\ +\ 5\right)}{x\left(3y\ +\ 4\right)\ -\ 1\left(3y\ +\ 4\right)})
And then factoring 
from the two terms that remain:
\left(2y\ +\ 5\right)}{\left(x\ -\ 1\right)\left(3y\ +\ 4\right)})
Eliminate the factor of 
that is common to both numerator and denominator leaving you with:
John
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