SOLUTION: solve. 3x/x^2-1=5/x I worked it out to: 2(x-5)(x+1) Is this correct?

Algebra ->  Equations -> SOLUTION: solve. 3x/x^2-1=5/x I worked it out to: 2(x-5)(x+1) Is this correct?      Log On


   

Question 191061This question is from textbook saxon algebra2
: solve.
3x/x^2-1=5/x
I worked it out to:
2(x-5)(x+1)
Is this correct? This question is from textbook saxon algebra2

Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that the equation is %283x%29%2F%28x%5E2-1%29=5%2Fx.


%283x%29%2F%28x%5E2-1%29=5%2Fx Start with the given equation


%283x%29%28x%29=5%28x%5E2-1%29 Cross multiply


3x%5E2=5x%5E2-5 Distribute and multiply


3x%5E2-5x%5E2=-5 Subtract 5x%5E2 from both sides


-2x%5E2=-5 Combine like terms.


x%5E2=-5%2F%28-2%29 Divide both sides by -2


x%5E2=5%2F2 Reduce


x=%22%22%2B-sqrt%285%2F2%29 Take the square root of both sides



x=sqrt%285%2F2%29 or x=-sqrt%285%2F2%29 Break up the "plus/minus" to form 2 separate equations.


x=sqrt%2810%29%2F2 or x=-sqrt%2810%29%2F2 Simplify the square root.


So the solutions are x=sqrt%2810%29%2F2 or x=-sqrt%2810%29%2F2

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Well, I will give you high marks for creativity. But your answer isn't even close.



The two denominators have no factors in common, so the LCD is simply the product of the two, or . Multiplying that out would only confuse the issue, so let's leave it the way it is. Applying the LCD:




Now, use the distributive property on the numerator on the right:



Now we have two fractions equal to each other and they have equal denominators. Therefore the numerators must be equal as well. So:



Add to both sides:



Multiply both sides by



Take the square root:



But we still need to rationalize the denominator so multiply by 1 in the form of



You should check both answers to make certain that we didn't create an extraneous root in the action of squaring the variable during the solving process.

John