SOLUTION: The whole rational equations has me stumped! Can you please help with this one? Solve... 3a-5/a^2+4a+3 + 2a+2/a+3 = a-3/a+1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The whole rational equations has me stumped! Can you please help with this one? Solve... 3a-5/a^2+4a+3 + 2a+2/a+3 = a-3/a+1      Log On


   



Question 239034: The whole rational equations has me stumped! Can you please help with this one?
Solve...
3a-5/a^2+4a+3 + 2a+2/a+3 = a-3/a+1

Answer by College Student(505) About Me  (Show Source):
You can put this solution on YOUR website!
Problem:

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It took me several steps to solve this problem. Look carefully as you move through the steps.
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First make eqn equal to zero:

Now factor first denominator:

I would also factor the second numerator. You'll know why in a moment... it'll make our lives easier.

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* In order to add fractions, we need common denominators, right?
* We also know that any number multiplied by 1 is equal to itself, in other words, it doesn't change the original value.
* Furthermore, we know that any number divided by itself is equal to 1, right?
* So, in order to have equal denominators, we need to multiply the second fraction by %28a%2B1%29%2F%28a%2B1%29 and the third denominator by %28a%2B3%29%2F%28a%2B3%29.
* This is how the new equation would look like:
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Now in a more cohesive way:

Since we now have the same denominator, we can group all numerators above one common denominator.

Now perform multiplications, leave denominator as is.

Distribute the negative sign in -%28a%5E2-9%29

Multiply the 2 in 2%28a%5E2%2B2a%2B1%29
%283a-5%2B2a%5E2%2B4a%2B2-a%5E2%2B9%29%2F%28%28a%2B1%29%28a%2B3%29%29=0
Combine like terms:
%283a%2B4a-5%2B2%2B9%2B2a%5E2-a%5E2%29%2F%28%28a%2B1%29%28a%2B3%29%29=0
%287a%2B6%2Ba%5E2%29%2F%28%28a%2B1%29%28a%2B3%29%29=0
Now in standard form to make it easier to factor:
%28a%5E2%2B7a%2B6%29%2F%28%28a%2B1%29%28a%2B3%29%29=0
%28%28a%2B6%29%28a%2B3%29%29%2F%28%28a%2B1%29%28a%2B3%29%29=0
Now simplify and you're done.
%28a%2B6%29%2F%28a%2B1%29
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It's a long problem. I hope you understood it and were able to follow it. :)