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Algebra 2 help please!? Simplify the sum. State any restrictions on the variables.

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Hint: Multiply the first fraction by (x - 3)/(x-3)
so would it become \[\frac{ x ^{2} + 5x + 6 }{ x ^{2} - 9 }\] @Hero

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Other answers:

  • phi
are you posting one of the multiple choices ?
This isn't a multiple choice problem
  • phi
you have the wrong sign on 5x
Actually the numerator should be x^2 - 5x + 6
Okay, then what would I do?
  • phi
once you have a common denominator, you can combine the tops
Since you are adding fractions, you need a common denominator. The denominator of the left fraction is simply x + 3. You need to factor the denominator of the right fraction. x^2 - 9 = (x + 3)(x - 3)
Since the right denominator has factors (x + 3)(x - 3) and the left fraction only has (x + 3), you need to multiply the numerator and denominator of the left fraction by (x - 3).
Okay, I get you so far
Then multiply out the numerator of the left fraction (x + 3)(x - 2)
Okay I did that
What did you do here?
Since both fractions now have the same denominator, you can write them as a single fraction over the common denominator. Now combine like terms in the numerator.
Ohh okay
I multiplied out the left numerator and I added the right numerator, and set the whole thing over the common denominator.
Now combine like terms on the numerator.
Then try factoring the numerator.
When I combine like terms I got \[\frac{ x ^{2} + 11x - 6 }{ (x+3)(x-3) }\]
Correct. The last step is to try to factor the numerator to see if you can simplify the fraction.
I can't figure out how to factro this, nothing I can find will come up with both 11 and 6
This kind of factoring involves finding two numbers that multiply to -6 and add to 11. There aren't any, so it can't be factored, and the addition is finished.
Okay, What would the restrictions be then ? @mathstudent55
Would the answer be with the denominator factored or can I put x^2 - 9 ?
The restrictions are any values of x that would make the denominator zero. Since the denominator is x^2 - 9 which you know factors into (x + 3)(x - 3), set x + 3 = 0 and solve for x and set x - 3 = 0 and solve for x. Those two x values are the restrictions.
Okay thank the answer woould be \[\frac{ x ^{2} + 11x - 6 }{ (x + 3)(x - 3) } ; x \neq -3, 3\] @mathstudent55
Or would I put x^2 - 9 as the denominator ?
You can leave the denominator factored. It's perfectly acceptable. It's also fine to multiply it out. Either way is good.
Okay, thank you so much for your help ! (:

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