Addition and Subtraction of  Rational Expressions

16 Addition and Subtraction of Rational Expressions

$\displaystyle{ \frac{P}{Q} + \frac{S}{T} = \frac{P T + Q S }{ST} }$

$\displaystyle{ \frac{P}{Q} - \frac{S}{T} = \frac{P T - Q S }{ST} }$

$\displaystyle{ \frac{P}{R} + \frac{S}{R} = \frac{P + S }{R} }$

$\displaystyle{ \frac{P}{R} - \frac{S}{R} = \frac{P - S }{R} }$

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Exercise 1

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We have

$\displaystyle{ x+4 + \frac{4}{ x+4 } = \frac{x+4}{1 } - \frac{4}{x+4 } }$

$\displaystyle{=\frac{(x+4) (x+4)}{x+4 } - \frac{ 4}{x+4 } = \frac{ x^2 + 2 (4 )x + (4)^2-{4} }{x+4} = \frac{ x^2 + 8 x + 20 }{x+4} }$

The fraction is irreducible since the discriminant of $x^2 + 8 x + 20$ is $8^2 - 4 (1)(20) <0$ .

Exercise 2

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$\displaystyle{\frac{ x }{ x^2+ 5x +6} -\frac{2}{ x + 2} - \frac{2}{ x + 3} =\frac{ x }{ (x + 2)(x + 3)} -\frac{2}{ x + 2} - \frac{2}{ x + 3} }$

$\displaystyle{=\frac{ x }{ (x + 2)(x + 3)} -\frac{2 (x+3) }{ (x + 2)(x+3)} - \frac{2 (x+2)}{ (x+3)(x+2)}}$

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

$\displaystyle{= \frac{-3x - 10 }{(x + 2)(x+3) } }$

Exercise 3

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$\displaystyle{\frac{ x }{x^2 + x - 20} - \frac{ 5}{ x^2 -11x +28} =\frac{ x }{ (x - 4)(x + 5)} - \frac{5}{ (x - 4)(x - 7) } }$

$\displaystyle{=\frac{ x (x-7) }{ (x - 4)(x + 5)(x-7)} - \frac{5(x + 5)}{ (x - 4)(x-7) (x + 5)} }$

$\displaystyle{= \frac{ x^2 - ( 7 + 5) x - (5) (5) }{ (x - 4)(x-7) (x + 5)} }$

$\displaystyle{= \frac{ x^2 - 12x -25 }{ (x - 4)(x-7) (x + 5)} }$

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