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x+6/x^2-3x=A/x+B/x-3 Find A-B
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Partial fractions madness!\[\rm {x + 6 \over x^2 - 3x} = {A \over x} + {B \over x - 3}\]
The first thing to do is this:\[\rm {x+6 \over x(x - 3)} = {A \over x} + {B \over x - 3}\]
Now, multiply by the common denominator which is \(\rm x(x - 3)\) (both sides).
\[\rm x + 6 = A(x - 3) + B(x)\]
Now, substitute \(\rm x = -3\) to get the value of \(\rm B\).
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And to get the value of \(\rm A\), substitute \(\rm x = 0\).
so B = 3 and A = -2?
A=-2 B=3
No...
Substitute \(\rm x = 3\) to get the value of \(\rm B\).
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Yes, so B = 3.
Yeah, A = -2 B = 3
okay thanks!!
-2 - 3 = -5
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