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# partial fractions

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There are values A and B such that (Bx-11) / (x^2-7x+10) = A/(x-2) - 3/(x-5).
Find A+B

May 2, 2022

#1
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We have the equation: $${{bx-11} \over {x^2-7x+10}} = {a \over {x -2}} - {3 \over {x-5}}$$

We begin by multiplying $$a \over {x-2}$$ by $${x - 5} \over {x - 5}$$. This gives us $${ax-5a } \over {x^2-7x+10}$$

We can then multiply $${3 \over{x-5}}$$ by $${ x-2} \over {x -2}$$, which gives us $${3x-6} \over {x^2-7x+10}$$

Note: We don't have to do anything to the other side of the equation because we were multiplying the equation by 1

We now disregard the denominator (because they're the same) and have the equation: $$bx-11 = ax-5a+3x-6$$

We can now combine like terms: $$bx-ax=-5a+5+3x$$

Now, we can factor out the x: $$x(b-a)=-5a+5+3x$$

Dividing by x, we have: $${b-a= {{-5a+5+3x} \over x}}$$

Adding 2a to both sides, we get: $$b+a={{-5a +5 +3x+2a} \over x}$$

We can then simplify further: $$b+a=\color{brown}\boxed{-3a+5+3x \over x}$$

May 2, 2022
#2
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Multiply $$x^2 - 7x + 10 = (x -2)(x - 5)$$ on both sides:

$$Bx - 11 = A(x - 5) - 3(x - 2)$$

Since this is true for any real number x, we can substitute whatever real number we want.

Substitute x = 5.

$$5B - 11 = -3(5 - 2)\\ 5B - 11 = -9\\ 5B = 2\\ B = \dfrac25$$

Substitute B = 2/5 and x = 2.

$$\dfrac45 - 11 = A(2 - 5)\\ A = \dfrac{17}5$$

May 2, 2022