Partial Fraction Decomposition Of (11-x)/((2x-1)(x+3))

In the realm of algebra, the technique of partial fraction decomposition stands as a powerful tool for simplifying complex rational expressions. At its core, this method allows us to break down a single fraction with a complicated denominator into a sum of simpler fractions, each having a denominator that is a factor of the original. This decomposition is incredibly useful in various mathematical contexts, including calculus (for integration), solving differential equations, and in areas of engineering and physics. Our focus is on expressing the rational function in a more manageable form that facilitates further analysis or computation.

The concept hinges on the idea that any rational function, where the degree of the numerator is less than the degree of the denominator, can be written as a sum of fractions whose denominators are the factors of the original denominator. The process involves setting up an equation where the original fraction is equated to the sum of the partial fractions, each with an unknown constant in the numerator. The key to success lies in determining these constants, which can be achieved through algebraic manipulation, such as equating coefficients or substituting specific values of the variable.

This technique is not just a mathematical trick; it is deeply rooted in the properties of polynomials and rational functions. Understanding partial fractions provides a deeper insight into the structure of algebraic expressions and their behavior. Furthermore, the ability to decompose fractions simplifies many complex problems, making it an indispensable tool in the arsenal of any mathematician, scientist, or engineer. The practical applications of partial fractions are vast and varied, underscoring its importance in both theoretical and applied mathematics. By mastering this technique, one gains a significant advantage in tackling a wide range of problems across different disciplines.

Our specific task is to express the rational function ${(11-x)/((2x-1)(x+3))}$ in partial fractions. This means we aim to rewrite this fraction as a sum of two simpler fractions, each with a denominator corresponding to one of the factors of the original denominator. The factors of the denominator are ${(2x-1)}$ and ${(x+3)}$. Thus, we anticipate our decomposition to take the form:

${ \frac{11-x}{(2x-1)(x+3)} = \frac{A}{2x-1} + \frac{B}{x+3} }$

where ${A}$ and ${B}$ are constants that we need to determine. The crux of the problem now lies in finding the values of these constants. There are several methods to achieve this, including the method of equating coefficients and the substitution method. Each approach offers a unique perspective and can be more efficient depending on the specific problem. Our goal is to find the most straightforward and accurate method to solve for ${A}$ and ${B}$. Once we have these values, we will have successfully decomposed the original fraction into its partial fractions.

This decomposition is not merely an exercise in algebraic manipulation; it is a fundamental step towards simplifying more complex problems. For instance, in calculus, integrating the original fraction directly can be challenging, but integrating its partial fraction decomposition is often much simpler. Similarly, in control systems engineering, partial fraction decomposition is used to analyze the stability and response of systems. Therefore, mastering this technique is crucial for anyone working in these fields. The problem at hand serves as a concrete example of how to apply the general theory of partial fractions to a specific case, providing a solid foundation for tackling more complex scenarios.

The method of equating coefficients is a powerful algebraic technique used to solve for unknown constants in an equation. In the context of partial fractions, this method involves clearing the denominators in our equation and then comparing the coefficients of like terms on both sides. This comparison results in a system of linear equations that we can solve to find the unknown constants.

Starting with our equation:

${ \frac{11-x}{(2x-1)(x+3)} = \frac{A}{2x-1} + \frac{B}{x+3} }$

We first clear the denominators by multiplying both sides by ${(2x-1)(x+3)}$. This gives us:

${ 11-x = A(x+3) + B(2x-1) }$

Next, we expand the right side of the equation:

${ 11-x = Ax + 3A + 2Bx - B }$

Now, we group the terms with the same powers of ${x}$:

${ 11-x = (A+2B)x + (3A-B) }$

For this equation to hold true for all values of ${x}$, the coefficients of the corresponding powers of ${x}$ on both sides must be equal. This gives us two equations:

1. Coefficient of ${x}$: ${A + 2B = -1}$
2. Constant term: ${3A - B = 11}$

We now have a system of two linear equations in two unknowns, which we can solve using various methods, such as substitution or elimination. Let's use the method of elimination. Multiply the second equation by 2 to make the coefficients of ${B}$ opposites:

1. ${A + 2B = -1}$
2. ${6A - 2B = 22}$

Adding the two equations, we get:

${ 7A = 21 }$

Dividing by 7, we find:

${ A = 3 }$

Now, substitute ${A = 3}$ into the first equation:

${ 3 + 2B = -1 }$

Subtract 3 from both sides:

${ 2B = -4 }$

Divide by 2:

${ B = -2 }$

Thus, we have found the values of the constants: ${A = 3}$ and ${B = -2}$. This method provides a systematic approach to solving for the unknown constants by leveraging the properties of polynomial equality.

The substitution method offers an alternative approach to finding the constants ${A}$ and ${B}$ in our partial fraction decomposition. This method involves substituting specific values of ${x}$ into the equation to eliminate one of the unknowns, allowing us to solve for the other. The key is to choose values of ${x}$ that make one of the factors in the equation equal to zero, thereby simplifying the equation.

Recall our equation after clearing the denominators:

${ 11-x = A(x+3) + B(2x-1) }$

We can choose values of ${x}$ that will make either ${(x+3)}$ or ${(2x-1)}$ equal to zero. First, let's choose ${x = -3}$. Substituting this into the equation, we get:

${ 11 - (-3) = A(-3+3) + B(2(-3)-1) }$

Simplifying, we have:

${ 14 = A(0) + B(-7) }$

${ 14 = -7B }$

Dividing by -7, we find:

${ B = -2 }$

Now, let's choose ${x}$ such that ${2x-1 = 0}$, which means ${x = \frac{1}{2}}$. Substituting this into the equation, we get:

${ 11 - \frac{1}{2} = A(\frac{1}{2}+3) + B(2(\frac{1}{2})-1) }$

Simplifying, we have:

${ \frac{21}{2} = A(\frac{7}{2}) + B(0) }$

${ \frac{21}{2} = \frac{7}{2}A }$

Multiplying both sides by ${\frac{2}{7}}$, we find:

${ A = 3 }$

Thus, we have found the values of the constants: ${A = 3}$ and ${B = -2}$. This method demonstrates the power of strategic substitution in simplifying algebraic equations. By choosing specific values of ${x}$, we can effectively isolate the unknowns and solve for them more easily. This approach is particularly useful when the factors in the denominator are linear, as it allows for straightforward elimination of terms.

Having found the constants ${A = 3}$ and ${B = -2}$ using both the equating coefficients and substitution methods, we can now express the original fraction in partial fractions:

${ \frac{11-x}{(2x-1)(x+3)} = \frac{3}{2x-1} + \frac{-2}{x+3} }$

Which can be written as:

${ \frac{11-x}{(2x-1)(x+3)} = \frac{3}{2x-1} - \frac{2}{x+3} }$

To verify our solution, we can combine the partial fractions back into a single fraction and see if it matches the original expression. To do this, we find a common denominator, which is ${(2x-1)(x+3)}$:

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

Now, we expand the numerators:

${ = \frac{3x+9}{(2x-1)(x+3)} - \frac{4x-2}{(2x-1)(x+3)} }$

Combine the fractions:

${ = \frac{(3x+9) - (4x-2)}{(2x-1)(x+3)} }$

Simplify the numerator:

${ = \frac{3x+9-4x+2}{(2x-1)(x+3)} }$

${ = \frac{-x+11}{(2x-1)(x+3)} }$

${ = \frac{11-x}{(2x-1)(x+3)} }$

Our result matches the original fraction, thus verifying our partial fraction decomposition. This verification step is crucial as it confirms the accuracy of our calculations and ensures that we have correctly decomposed the fraction.

In conclusion, we have successfully expressed the rational function ${(11-x)/((2x-1)(x+3))}$ in partial fractions as:

${ \frac{11-x}{(2x-1)(x+3)} = \frac{3}{2x-1} - \frac{2}{x+3} }$

We achieved this decomposition using two methods: the method of equating coefficients and the substitution method. Both methods provided the same result, showcasing the versatility and robustness of these algebraic techniques. The verification step further solidified our solution, ensuring its accuracy.

The significance of partial fraction decomposition extends far beyond this specific example. As mentioned earlier, this technique is invaluable in calculus, particularly in integration. Many integrals involving rational functions become significantly easier to solve once the function is decomposed into partial fractions. Similarly, in differential equations, partial fractions are used to simplify the process of finding solutions, especially when dealing with Laplace transforms.

Furthermore, partial fractions play a crucial role in engineering and physics. In electrical engineering, they are used to analyze circuits and systems, while in control systems, they help in determining system stability and response. The ability to break down complex rational functions into simpler components allows engineers and physicists to model and understand complex phenomena more effectively.

In essence, partial fraction decomposition is a fundamental tool in mathematics and its applications. It provides a systematic way to simplify rational functions, making them more amenable to analysis and computation. By mastering this technique, one gains a powerful tool for tackling a wide range of problems in various fields. The process not only enhances algebraic skills but also deepens the understanding of the structure and behavior of rational functions.