Problem 12
Question
Find the partial fraction decomposition for each rational expression. $$\frac{5 x-3}{(x+1)(x-3)}$$
Step-by-Step Solution
Verified Answer
The partial fraction decomposition is \( \frac{2}{x+1} + \frac{3}{x-3} \).
1Step 1: Identify the Form of Decomposition
The given expression is \( \frac{5x-3}{(x+1)(x-3)} \). Since both \(x+1\) and \(x-3\) are linear factors, express the decomposition as \( \frac{A}{x+1} + \frac{B}{x-3} \).
2Step 2: Write Equation from Decomposition
Set the expression equal to its decomposition: \( \frac{5x-3}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3} \).
3Step 3: Clear the Denominator
Multiply both sides by \((x+1)(x-3)\) to get rid of the denominator: \[ 5x-3 = A(x-3) + B(x+1) \]
4Step 4: Expand and Simplify
Expand the right side: \[ 5x-3 = Ax - 3A + Bx + B \] Combine like terms: \[ 5x-3 = (A+B)x + (-3A + B) \]
5Step 5: Equate Coefficients
Compare both sides to form two equations: - Coefficient of \(x\): \( A + B = 5 \) - Constant term: \( -3A + B = -3 \)
6Step 6: Solve the System of Equations
Solve these equations simultaneously: 1. \( A + B = 5 \) 2. \( -3A + B = -3 \)Subtract equation 1 from equation 2:\(-3A + B - (A + B) = -3 - 5\)\(-4A = -8\)Therefore, \( A = 2 \).Substitute \( A = 2 \) back into equation 1: \( 2 + B = 5 \) gives \( B = 3 \).
7Step 7: Write the Partial Fraction Decomposition
Substitute \( A \) and \( B \) back into the decomposition: \[ \frac{5x-3}{(x+1)(x-3)} = \frac{2}{x+1} + \frac{3}{x-3} \].
Key Concepts
Rational ExpressionLinear FactorsSystem of Equations
Rational Expression
A rational expression is essentially a fraction where both the numerator and the denominator are polynomials. In our exercise, the expression \( \frac{5x-3}{(x+1)(x-3)} \) is a rational expression. The numerator is the polynomial \( 5x-3 \), while the denominator is \( (x+1)(x-3) \).
Rational expressions are significantly important in calculus and algebra. They help in simplifying complex expressions and solving equations. Like regular fractions, you can add, subtract, multiply, and divide rational expressions, but you must first find a common denominator for addition and subtraction.
Partial fraction decomposition is a technique used to express a rational expression as the sum of simpler fractions with linear or quadratic denominators. This method is particularly useful in integration because it allows us to deal with simpler terms one at a time.
To decompose a rational expression, we first identify the type of factors in the denominator. In our problem, both \( x+1 \) and \( x-3 \) are linear, making our decomposition form straightforward.
Rational expressions are significantly important in calculus and algebra. They help in simplifying complex expressions and solving equations. Like regular fractions, you can add, subtract, multiply, and divide rational expressions, but you must first find a common denominator for addition and subtraction.
Partial fraction decomposition is a technique used to express a rational expression as the sum of simpler fractions with linear or quadratic denominators. This method is particularly useful in integration because it allows us to deal with simpler terms one at a time.
To decompose a rational expression, we first identify the type of factors in the denominator. In our problem, both \( x+1 \) and \( x-3 \) are linear, making our decomposition form straightforward.
Linear Factors
In the context of rational expressions, a linear factor in the denominator can appear in simple forms like \( (x + a) \). For the expression \( \frac{5x-3}{(x+1)(x-3)} \), both \( x+1 \) and \( x-3 \) are linear factors.
A linear factor is one where the variable \( x \) is raised to the first power. This characteristic makes dealing with linear factors more manageable when decomposing into partial fractions. Each linear factor corresponds to a separate term in the decomposition. That is why we express the rational expression as \( \frac{A}{x+1} + \frac{B}{x-3} \), where \( A \) and \( B \) are constants that we will determine.
It's crucial to correctly identify these linear factors early in the process. During decomposition, each factor in the denominator will result in a separate fraction in the decomposition. This setup allows us to break down complex fractions into simpler, more manageable pieces.
A linear factor is one where the variable \( x \) is raised to the first power. This characteristic makes dealing with linear factors more manageable when decomposing into partial fractions. Each linear factor corresponds to a separate term in the decomposition. That is why we express the rational expression as \( \frac{A}{x+1} + \frac{B}{x-3} \), where \( A \) and \( B \) are constants that we will determine.
It's crucial to correctly identify these linear factors early in the process. During decomposition, each factor in the denominator will result in a separate fraction in the decomposition. This setup allows us to break down complex fractions into simpler, more manageable pieces.
System of Equations
A system of equations is a set of equations with the same variables. To find specific values that satisfy all equations in the system, we must solve them simultaneously.
In partial fraction decomposition, we often set up a system of equations to find unknown coefficients such as \( A \) and \( B \). From our problem, after equating the original expression and its decomposition form, we transformed it into:
- Coefficient of \(x\): \( A + B = 5 \)
- Constant term: \( -3A + B = -3 \)
We then solve this system using methods like substitution or elimination. Solving these gives the specific values of \( A \) and \( B \) that make the decomposition valid. For our problem, we found \( A = 2 \) and \( B = 3 \).
Understanding how to solve a system of equations effectively is pivotal, as this process often surfaces in algebra, calculus, and beyond. Look for patterns in equations, and remember that the goal is to find values that work for all given equations simultaneously.
In partial fraction decomposition, we often set up a system of equations to find unknown coefficients such as \( A \) and \( B \). From our problem, after equating the original expression and its decomposition form, we transformed it into:
- Coefficient of \(x\): \( A + B = 5 \)
- Constant term: \( -3A + B = -3 \)
We then solve this system using methods like substitution or elimination. Solving these gives the specific values of \( A \) and \( B \) that make the decomposition valid. For our problem, we found \( A = 2 \) and \( B = 3 \).
Understanding how to solve a system of equations effectively is pivotal, as this process often surfaces in algebra, calculus, and beyond. Look for patterns in equations, and remember that the goal is to find values that work for all given equations simultaneously.
Other exercises in this chapter
Problem 12
Graph each inequality. $$x-y \geq 2$$
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For each element in the second row of the given matrix, find its cofactor (See Example 3 .) $$\left[\begin{array}{rrr}2 & -1 & 4 \\\3 & 0 & 1 \\\\-2 & 1 & 4\end
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For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{rr} -1 & 2 \\ -2 & -1 \end{array}\right]$$
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Write the augmented matrix for each system. Do not solve the system. $$\begin{aligned} &4 x-2 y+3 z=4\\\ &3 x+5 y+z=7\\\ &5 x-y+4 z=7 \end{aligned}$$
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