Add: $\frac{4x+3}{x^2-9} - \frac{2x}{3-x}$

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Introduction

Complex fractions can be a daunting task for many students and mathematicians alike. However, with the right approach and techniques, simplifying these fractions can become a manageable and even enjoyable process. In this article, we will explore the process of simplifying complex fractions, focusing on the specific example of adding two fractions: 4x+3x2βˆ’9βˆ’2x3βˆ’x\frac{4x+3}{x^2-9} - \frac{2x}{3-x}.

Understanding Complex Fractions

A complex fraction is a fraction that contains one or more fractions in its numerator or denominator. In the given example, both fractions have complex numerators and denominators, making it a complex fraction. To simplify complex fractions, we need to follow a step-by-step approach that involves factoring, canceling, and combining like terms.

Step 1: Factor the Denominators

The first step in simplifying complex fractions is to factor the denominators. In the given example, we have two fractions with the following denominators:

  • x2βˆ’9x^2-9
  • 3βˆ’x3-x

We can factor the first denominator as (x+3)(xβˆ’3)(x+3)(x-3) and the second denominator as (3βˆ’x)(3-x) or βˆ’(xβˆ’3)-(x-3).

Step 2: Rewrite the Fractions with Factored Denominators

Now that we have factored the denominators, we can rewrite the fractions with the factored denominators:

  • 4x+3(x+3)(xβˆ’3)\frac{4x+3}{(x+3)(x-3)}
  • 2xβˆ’(xβˆ’3)\frac{2x}{-(x-3)}

Step 3: Simplify the Fractions

To simplify the fractions, we need to cancel out any common factors between the numerators and denominators. In the first fraction, we can cancel out the common factor of (xβˆ’3)(x-3) between the numerator and denominator:

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

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Simplifying Complex Fractions: A Step-by-Step Guide

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Understanding Complex Fractions

A complex fraction is a fraction that contains one or more fractions in its numerator or denominator. In the given example, both fractions have complex numerators and denominators, making it a complex fraction. To simplify complex fractions, we need to follow a step-by-step approach that involves factoring, canceling, and combining like terms.

Step 1: Factor the Denominators

The first step in simplifying complex fractions is to factor the denominators. In the given example, we have two fractions with the following denominators:

  • x2βˆ’9x^2-9
  • 3βˆ’x3-x

We can factor the first denominator as (x+3)(xβˆ’3)(x+3)(x-3) and the second denominator as (3βˆ’x)(3-x) or βˆ’(xβˆ’3)-(x-3).

Step 2: Rewrite the Fractions with Factored Denominators

Now that we have factored the denominators, we can rewrite the fractions with the factored denominators:

  • 4x+3(x+3)(xβˆ’3)\frac{4x+3}{(x+3)(x-3)}
  • 2xβˆ’(xβˆ’3)\frac{2x}{-(x-3)}

Step 3: Simplify the Fractions

To simplify the fractions, we need to cancel out any common factors between the numerators and denominators. In the first fraction, we can cancel out the common factor of (xβˆ’3)(x-3) between the numerator and denominator:

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.

4x+3(x+3)(xβˆ’3)=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3=4x+3(x+3)(xβˆ’3)β‹…1xβˆ’3\frac{4x+3}{(x+3)(x-3)} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3} = \frac{4x+3}{(x+3)(x-3)} \cdot \frac{1}{x-3}

However, we can simplify it by cancelling the (x-3) from the numerator and denominator.