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Answer:

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Step-by-step explanation:

Given

[tex]\frac{2x-1}{x^2-2x-3}[/tex] + [tex]\frac{1}{x-3}[/tex]

Factorise the denominator of the first fraction

x² - 2x - 3 = (x - 3)(x + 1), then

= [tex]\frac{2x-1}{(x-3)(x+1)}[/tex] + [tex]\frac{1}{x-3}[/tex]

To add we require the fractions to have a common denominator.

Multiply the numerator/denominator of the fraction on the right by (x + 1)

= [tex]\frac{2x-1}{(x-3)(x+1)}[/tex] + [tex]\frac{x+1}{(x-3)(x+1)}[/tex]

Now add the terms on the numerators leaving the denominator

= [tex]\frac{2x-1+x+1}{(x-3)(x+1)}[/tex]

= [tex]\frac{3x}{(x-3)(x+1)}[/tex]

akposevictor

In order to simplify  [tex]\frac{2x - 1}{x^2 - 2x - 3} + \frac{1}{x - 3}[/tex] , we would factorize its denominator,  [tex]x^2 - 2x - 3[/tex] , then simplify further to get our final answer as: [tex]\frac{3x}{(x - 3)(x +1)}[/tex]

To simplify, [tex]\frac{2x - 1}{x^2 - 2x - 3} + \frac{1}{x - 3}[/tex], first we would need to factorize [tex]x^2 - 2x - 3[/tex], which is the denominator of the first fraction.

  • Thus:

[tex]\frac{2x - 1}{x^2 + x -3x - 3} + \frac{1}{x - 3}\\\\\frac{2x - 1}{x(x + 1)-3(x +1)} + \frac{1}{x - 3}\\\\\frac{2x - 1}{(x - 3)(x +1)} + \frac{1}{x - 3}[/tex]

To make both fractions as one:

  • Find the common denominator of both fractions (common denominator would be, (x - 3)(x + 1).
  • Divide the denominator of each of the fractions and multiply what you get by their numerator to make both fractions as one.

  • Thus:

[tex]\frac{(1)(2x - 1) + (x + 1)(1)}{(x - 3)(x +1)} \\\\\frac{2x - 1 + x + 1}{(x - 3)(x +1)}\\\\[/tex]

  • Simplify further

[tex]\frac{2x + x - 1 + 1}{(x - 3)(x +1)}\\\\\frac{3x}{(x - 3)(x +1)}\\\\[/tex]

Therefore, in order to simplify  [tex]\frac{2x - 1}{x^2 - 2x - 3} + \frac{1}{x - 3}[/tex] , we would factorize its denominator,  [tex]x^2 - 2x - 3[/tex] , then simplify further to get our final answer as: [tex]\frac{3x}{(x - 3)(x +1)}[/tex]

Learn more here:

https://brainly.com/question/21191537

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