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FinnZ [79.3K]
4 years ago
13

Am I doing this math correctly?

Mathematics
1 answer:
11Alexandr11 [23.1K]4 years ago
7 0
Yes,look good,your math is correct :D
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Given that (7x-8)/(2x-1)(x-2) <br><br> Use Partial Fraction Decomposition to Find A and B.
koban [17]

We look for constants <em>a</em> and <em>b</em> such that

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac a{2x-1} + \frac b{x-2}

Rewrite all terms with a common denominator and set the numerators equal:

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)}{(2x-1)(x-2)} + \frac{b(2x-1)}{(x-2)(2x-1)} \\\\ \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)+b(2x-1)}{(2x-1)(x-2)} \\\\ \implies 7x-8 = a(x-2) + b(2x-1) = (a+2b)x - 2a - b

Then

<em>a</em> + 2<em>b</em> = 7

-2<em>a</em> - <em>b</em> = -8

Solve for <em>a</em> and <em>b</em>. Using elimination: multiply the first equation by 2 and add it to the second equation:

2 (<em>a</em> + 2<em>b</em>) + (-2<em>a</em> - <em>b</em>) = 2(7) + (-8)

2<em>a</em> + 4<em>b</em> - 2<em>a</em> - <em>b</em> = 14 - 8

3<em>b</em> = 6

<em>b</em> = 2

Then

<em>a</em> + 2(2) = 7   ==>   <em>a</em> = 3

and so

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac 3{2x-1} + \frac 2{x-2}

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