Rational equation: (2x-9)/(x^2-x-6)=A/(x-3)+B/(x+2)

MaxRabbit

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Hello-not sure about this problem; I need to find constants A and B that make the equation true. The equation is as follows:

(2x-9) over (x^2-x-6)=(A) over (x-3)+(B) over (x+2)

Could you please, in steps, help me to solve this equation?
 
Re: Rational Expression equation

Simply add the fractions on the right hand side and you should see how to proceed.
 
Re: Rational Expression equation

MaxRabbit said:
Hello-not sure about this problem; I need to find constants A and B that make the equation true. The equation is as follows:

(2x-9) over (x^2-x-6)=(A) over (x-3)+(B) over (x+2)

Could you please, in steps, help me to solve this equation?

\(\displaystyle \frac{2x-9}{(x-3)(x+2)} \, = \, \frac{A}{x-3} \, + \, \frac{B}{x+2}\)

multiply both sides by (x-3)

\(\displaystyle \frac{2x-9}{(x+2)} \, = \, A} \, + \, \frac{B(x-3)}{x+2}\)

Now let x = 3

A = 3

Similarly find B
 
Re: Rational Expression equation

I don't understand how you figured out to let x=3?

Edit: Never mind-I get it; you let x=3 in order to eliminate B and solve for A!
 
Re: Rational Expression equation

Subhotosh Khan said:
\(\displaystyle \frac{2x-9}{(x-3)(x+2)} \, = \, \frac{A}{x-3} \, + \, \frac{B}{x+2}\)

multiply both sides by (x-3)

\(\displaystyle \frac{2x-9}{(x+2)} \, = \, A} \, + \, \frac{B(x-3)}{x+2}\)

Now let x = 3

A = 3

Similarly find B
Don't get it :( Once you make x = 3, then you come up with -3/5 = A.....?
 
Re: Rational Expression equation

\(\displaystyle \frac{2x-9}{x+2}\, = \,\frac{2\cdot 3-9}{3+2}\, = -\frac{3}{5}\)

Did you use pencil/paper - or just stared at the screen?
 
Re: Rational Expression equation

Subhotosh Khan said:
\(\displaystyle \frac{2x-9}{x+2}\, = \,\frac{2\cdot 3-9}{3+2}\, = -\frac{3}{5}\)

Did you use pencil/paper - or just stared at the screen?
Pencil and paper-why? Am I missing something stupid?
 
So, you're saying you can't just add the fractions?

\(\displaystyle \frac{A}{x-3}+\frac{B}{x+2}\;=\;\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}\;=\;\frac{(A+B)x + (2A-3B)}{(x-3)(x+2)}\)

I have to believe that's useful.
 
tkhunny said:
So, you're saying you can't just add the fractions?

\(\displaystyle \frac{A}{x-3}+\frac{B}{x+2}\;=\;\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}\;=\;\frac{(A+B)x + (2A-3B)}{(x-3)(x+2)}\)

I have to believe that's useful.
Idk really what to do with that, though...
 
MaxRabbit said:
Idk really what to do with that, though...
I think the above means that you aren't familiar with how to work with polynomial fractions...? (I'm not entirely familiar with the cutesy chat-speak lingo, but I'm guessing that "Idk" is meant to by "I don't know". My apologies if I've guessed your meaning incorrectly.)

Are you familiar with how to add numerical fractions? Or should we start on that topic, before moving on to polynomial fractions?

Thank you! :D

Eliz.
 
Re:

stapel said:
MaxRabbit said:
Idk really what to do with that, though...
I think the above means that you aren't familiar with how to work with polynomial fractions...? (I'm not entirely familiar with the cutesy chat-speak lingo, but I'm guessing that "Idk" is meant to by "I don't know". My apologies if I've guessed your meaning incorrectly.)

Are you familiar with how to add numerical fractions? Or should we start on that topic, before moving on to polynomial fractions?

Thank you! :D

Eliz.
Yes idk is I don't know :) And I perfectly understand adding polynomials-the problem is people say A = 3 but I get A = -3/5; Here is how I see it:
\(\displaystyle \text{We have: }\;\frac{2x-9}{(x-3)(x+2)} \;=\;\frac{A}{x-3} + \frac{B}{x+2}\)


\(\displaystyle \text{Multiply by the LCD: }\;2x - 9 \;=\;(x+2)A + (x-3)B\)

. . \(\displaystyle \begin{array}{cccccccc}\text{Let }x = 3: & \text{-}3 \;=\;5\!\cdot\!A + 0\!\cdot\!B & \Rightarrow& \boxed{A \;=\;\text{-}\frac{3}{5}} \\ \\ \text{Let }x = \text{-}2: & \text{-}13 \; = \; 0\!\cdot\!A + (\text{-}5)B & \Rightarrow & \boxed{B \;=\; \frac{13}{5}} \end{array}\)


\(\displaystyle \text{Therefore: }\;\frac{2x-9}{(x-3)(x+2)} \;=\;\frac{-\frac{3}{5}}{x-3} + \frac{\frac{13}{5}}{x+2}\)
 
MaxRabbit said:
...the problem is people say A = 3 but I get A = -3/5...
Which "people" are telling you that A equals 3...?

Note: The previous replies seem to agree with your value for A:

Subhotosh Khan said:
\(\displaystyle \frac{2x-9}{x+2}\, = \,\frac{2\cdot 3-9}{3+2}\, = -\frac{3}{5}\)
If you doubt your answer, try adding your two fractions together, and see if, after simplifying, you end up where you started. If you do, your decomposition is probably correct! :D

Eliz.
 
Look at the following graphic. It took less that 20 seconds to enter the code to get this result. The computer algebra system that did the work cost less that $40US.
Why in the world are we still putting students through the ordeal ‘partial fractions’? If any of us truly thinks that mathematics is about the possible description of reality, why not use every option open to us to get the ideas across
 

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One of my favorite methods of teaching during the early years of hand-held calculators was to take the calculator away and ask simply, "Now what?" It was not just a favorite method, it was also very successful in helping students think, rather than lose themselves in a technological device.

For those who go on, won't we need this idea for Complex Variables?

Until you win this argument, against an entire generation of local teachers of mathematics, given some partial fractions, shouldn't we learn to deal with them?

Note: I do not have a substantial rebuttal against your objection. I realize I was a freak in high school, one of very few who was motivated by meticulous rigor and personal concentration - avoiding outside influences of nearly any kind. If that's where we got all of our Math Majors, there would be many fewer of the breed.
 
Oh gee...you've hit on my "nerves" here....(maybe the very LAST of my nerves!). I agree that calculators and computers are very important tools these days...BUT...I'd surely love to think that students these days are using their BRAINS, and perhaps using computers and calculators to CHECK what they've thought about. I'm so disturbed by folks who think that if they can't get an answer on a computer or calculator, they're wrong. Back in the "old days," when one had to verify a concept "by hand," it was so gratifying when things worked.
 
Re: Re:

Max, didn't you know that rabbits multiply quickly?
 
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