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Linear Denominators Simplified Revision Notes

Revision notes with simplified explanations to understand Linear Denominators quickly and effectively.

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2.11.1 Linear Denominators

"Linear denominators" refer to algebraic fractions where the denominator is a linear expression. A linear expression is a polynomial of degree 11, typically in the form  ax+b,\ ax + b , where  a\ a and  b\ b are constants. Understanding how to work with linear denominators is important, especially in simplifying, adding, subtracting, and solving equations involving fractions.

Key Concepts

1. Simplifying Fractions with Linear Denominators

Simplifying a fraction involves finding the greatest common factor (GCF) between the numerator and the denominator and dividing both by this factor.

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Example: Simplify the fraction 2x+6x+3. \frac{2x + 6}{x + 3} .

Solution:

  • Factor the numerator:  2x+6=2(x+3).\ 2x + 6 = 2(x + 3) .
  • The fraction becomes  2(x+3)x+3.\ \frac{2(x + 3)}{x + 3} .
  • Since  x+3\ x + 3 is common to both the numerator and denominator, cancel it out: 2(x+3)x+3=2\frac{2(x + 3)}{x + 3} = 2 So, the simplified fraction is  2\ 2 .

2. Adding and Subtracting Fractions with Linear Denominators

To add or subtract fractions with different denominators, you first need to find a common denominator.

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Example: Add 1x+2 and 3x1. \frac{1}{x + 2} \ and \ \frac{3}{x - 1} .


Solution:

  1. Find the common denominator:  (x+2)(x1).\ (x + 2)(x - 1) .
  2. Rewrite each fraction with the common denominator: 1x+2=x1(x+2)(x1),3x1=3(x+2)(x+2)(x1)\frac{1}{x + 2} = \frac{x - 1}{(x + 2)(x - 1)}, \quad \frac{3}{x - 1} = \frac{3(x + 2)}{(x + 2)(x - 1)}
  3. Add the fractions: x1(x+2)(x1)+3(x+2)(x+2)(x1)=x1+3(x+2)(x+2)(x1)\frac{x - 1}{(x + 2)(x - 1)} + \frac{3(x + 2)}{(x + 2)(x - 1)} = \frac{x - 1 + 3(x + 2)}{(x + 2)(x - 1)}
  4. Simplify the numerator: x1+3(x+2)=x1+3x+6=4x+5x - 1 + 3(x + 2) = x - 1 + 3x + 6 = 4x + 5 So, 1x+2+3x1=4x+5(x+2)(x1).\text{So, } \frac{1}{x + 2} + \frac{3}{x - 1} = \frac{4x + 5}{(x + 2)(x - 1)}.

3. Solving Equations Involving Linear Denominators

To solve equations with linear denominators, you often need to eliminate the fractions by multiplying both sides of the equation by the least common denominator (LCD).

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Example: Solve  2x+1+3x2=1.\ \frac{2}{x + 1} + \frac{3}{x - 2} = 1 .


Solution:

  1. Find the common denominator:  (x+1)(x2).\ (x + 1)(x - 2) .
  2. Multiply each term by the common denominator to eliminate the fractions: 2(x2)x+1+3(x+1)x2=(x+1)(x2)\frac{2(x - 2)}{x + 1} + \frac{3(x + 1)}{x - 2} = (x + 1)(x - 2) This simplifies to: 2(x2)+3(x+1)=(x+1)(x2)2(x - 2) + 3(x + 1) = (x + 1)(x - 2)
  3. Expand and simplify: 2x4+3x+3=x2x22x - 4 + 3x + 3 = x^2 - x - 2 Combine like terms: 5x1=x2x25x - 1 = x^2 - x - 2
  4. Rearrange to form a quadratic equation: 0=x26x10 = x^2 - 6x - 1
  5. Solve using the quadratic formula: x=(6)±(6)24(1)(1)2(1)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2(1)} Simplify: x=6±36+42=6±402=6±2102=3±10x = \frac{6 \pm \sqrt{36 + 4}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10} So, the solutions are  x=3+10 and x=310\ x = 3 + \sqrt{10} \ and \ x = 3 - \sqrt{10} .

Practice Problem:

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Solve the equation 1x+22x3=0. \frac{1}{x + 2} - \frac{2}{x - 3} = 0 .

Solution:

  1. Multiply through by the common denominator  (x+2)(x3)\ (x + 2)(x - 3) to eliminate the fractions.
  2. Expand and simplify the resulting equation.
  3. Solve the resulting linear equation. This will give you the value of  x\ x .
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