How do you express $(5x - 4 ) / (x^2 -4x)$ in partial fractions?

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Answer 1 · 2016-10-22T20:41:58

$(5x - 4)/(x^2 - 4x) = 4/(x - 4) + 1/x$

The denominator can be written as $x(x - 4)$ .

We can rewrite this as follows:

$A/(x - 4) + B/x = (5x- 4)/(x^2 - 4x)$

Put on a common denominator.

$A(x) + B(x - 4) = 5x - 4$

$Ax + Bx - 4B = 5x - 4$

$(A + B)x - 4B = 5x - 4$

We can hence write the following system of equations.

$A + B = 5$
$-4B = -4$

Solving, we have that $B = 1$ and that $A = 4$

We can now reinsert into the original equation to get our partial fraction decomposition.

$4/(x- 4) + 1/x = (5x - 4)/(x^2 - 4x)$

Let's quickly check the validity of our answer.

What is the sum of $4/(x - 4) + 1/x$ ?

Put on a common denominator.

$(4(x))/((x - 4)(x)) + (1(x- 4))/(x(x - 4)) = (4x + x - 4)/(x(x - 4)) = (5x - 4)/(x^2 - 4x)$

This verifies the initial expression, so we have done our decomposition properly.

Hopefully this helps!

How do you express (5x - 4 ) / (x^2 -4x) (5x - 4 ) / (x^2 -4x) in partial fractions?