Express (9x-5)/(2(2x-1)(x-1)) in partial fractions.

Rewrite and Solve: Rewrite the expression as A/(2x-1) + B/(x-1) and solve for A and B.Clear Fractions: Multiply both sides by the denominator 2(2x-1)(x-1) to clear the fractions: (9x-5) = A(2(x-1)) + B(2x-1).Choose Values: Choose convenient values for x to solve for A and B. Let x=1, then (9*1-5) = B(2*1-1), so 4 = B(1), hence B = 4.Solve for A and B: Now let x=1/2, then (9*(1/2)-5) = A(2*(1/2-1)), so -4.5 = A(2*(-1/2)), hence -4.5 = -A, so A = 4.5.

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Algebra 2

Sum of finite series starts from 1

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Q. Express

\frac{9x-5}{2(2x-1)(x-1)}

in partial fractions.

Rewrite and Solve: Rewrite the expression as $\frac{A}{2x-1} + \frac{B}{x-1}$ and solve for $A$ and $B$ .
Clear Fractions: Multiply both sides by the denominator $2(2x-1)(x-1)$ to clear the fractions: $(9x-5) = A(2(x-1)) + B(2x-1)$ .
Choose Values: Choose convenient values for $x$ to solve for $A$ and $B$ . Let $x=1$ , then $(9\times 1-5) = B(2\times 1-1)$ , so $4 = B(1)$ , hence $B = 4$ .
Solve for $A$ and $B$ : Now let $x=\frac{1}{2}$ , then $(9\cdot\frac{1}{2}-5) = A(2\cdot(\frac{1}{2}-1))$ , so $-4.5 = A(2\cdot(-\frac{1}{2}))$ , hence $-4.5 = -A$ , so $A = 4.5$ .