ParveshD.Singh
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How do I simplify this into the following, I need to do this to solve a limit
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Harry_the_cat
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Can you do long division of poynomials?
Harry_the_cat
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Alternatively:
Let \(\displaystyle \frac{2x-3}{3x+4} = A + \frac{B}{3x+4}\)
\(\displaystyle \frac{2x-3}{3x+4} = \frac{A(3x+4)+B}{3x+4}\)
\(\displaystyle \frac{2x-3}{3x+4} = \frac{3Ax + (4A+B)}{3x+4}\)
then equate
3A=2 and 4A+B=-3 and solve for A and B using simultaneous equations.
Let \(\displaystyle \frac{2x-3}{3x+4} = A + \frac{B}{3x+4}\)
\(\displaystyle \frac{2x-3}{3x+4} = \frac{A(3x+4)+B}{3x+4}\)
\(\displaystyle \frac{2x-3}{3x+4} = \frac{3Ax + (4A+B)}{3x+4}\)
then equate
3A=2 and 4A+B=-3 and solve for A and B using simultaneous equations.
Last edited:
Dr.Peterson
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In this particular case, the ratio of leading coefficients is 2/3, giving you the desired constant term, and you can find the other by subtracting [MATH]\frac{2x-3}{3x+4} - \frac{2}{3}[/MATH] using the common denominator [MATH]3(3x+4)[/MATH]:
[MATH]\frac{2x-3}{3x+4} - \frac{2}{3}= \frac{3(2x-3)}{3(3x+4)} -\frac{2(3x+4)}{3(3x+4)} =\frac{6x-9}{3(3x+4)} -\frac{6x+8}{3(3x+4)}=\frac{-17}{3(3x+4)}[/MATH]
