Problem 5 - College Algebra

Question

Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had \(\frac{7 x+13}{3 x^{2}+8 x+15}=\frac{A}{x+1}+\frac{B}{3 x+5},\) we eventually simplify to \(7 x+13=A(3 x+5)+B(x+1)\) Explain how you could intelligently choose an \(x\) -value that will eliminate either \(A\) or \(B\) and solve for \(A\) and \(B\).

Step-by-Step Solution

Answer

Use strategic x-values to eliminate variables and solve.

Step 1: Initial Break-Down

We have the equation \( 7x + 13 = A(3x + 5) + B(x + 1) \). Our goal is to find the values of \( A \) and \( B \) by intelligently choosing values for \( x \) that simplify the equation.

Step 2: Choose a Strategic Value for x

First, select an \(x\)-value that will eliminate one of the variables. Set \( x = -1 \). This eliminates \( B \) because it results in \( B(x + 1) = B(0) \).

Step 3: Substitute x = -1 and Solve for A

Substitute \( x = -1 \) into the equation: \( 7(-1) + 13 = A(3(-1) + 5) + B(-1 + 1) \). This simplifies to \( -7 + 13 = A(2) + 0 \), leading to \( 6 = 2A \). Thus, \( A = 3 \).

Step 4: Choose a Second Strategic Value for x

Next, pick an \(x\)-value that will eliminate \( A \). Set \( x = -\frac{5}{3} \). This eliminates \( A \) because it results in \( A(3(-\frac{5}{3}) + 5) = A(0) \).

Step 5: Substitute x = -\( \frac{5}{3} \) and Solve for B

Substitute \( x = -\frac{5}{3} \) into the equation: \( 7(-\frac{5}{3}) + 13 = A(0) + B(-\frac{5}{3} + 1) \). Simplify to \( -\frac{35}{3} + 13 = B(-\frac{2}{3}) \). Convert 13 to \( \frac{39}{3} \) and simplify: \( \frac{4}{3} = -\frac{2}{3}B \). Solve to get \( B = -2 \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In our case, the expression from partial fraction decomposition results in the system of equations obtained by equating coefficients. In a system, each equation shares a common goal: solving for the variables simultaneously.
This concept often involves different variables interacting with each other in equations, making it important to find a solution that satisfies all equations involved. You can visualize this process as finding a common point or solution that exists within all the equations displayed in a multivariable space.
By using approaches like substitution or elimination, we aim to simplify the system until the value of each variable can be easily determined.

Strategic Substitution

Strategic substitution is a thoughtful approach to solve a system of equations where specific values for variables are carefully selected. In the context of balancing multiple equations such as those found from partial fraction decomposition, identifying particular values for a variable can make calculations simpler.
For example, if certain terms vanish when a specific variable value is substituted, it can greatly simplify the equation. This makes it easier to solve for other variables. In our exercise, choosing values like \( x = -1 \) or \( x = -\frac{5}{3} \) allows us to eliminate either A or B.
Strategically selecting such values means observing the structure of the equation and predicting how various substitutions impact the ease of calculating remaining variables.

Solving for Variables

The ultimate goal in working with systems of equations is to find the values of the variables, satisfying each equation. Solving for variables means isolating each variable on one side of the equation one at a time.
We start with substitutions or simplifications uniquely chosen to eliminate one of the variables, reducing the equation to a simpler form only involving one variable.

• Step 1: Identify a strategic substitution, like setting \( x = -1 \), which simplifies to find \( A \).
• Step 2: Use another strategic substitution, such as \( x = -\frac{5}{3} \), to solve for \( B \).

With these steps, each substitution reduces the complexity of the equation, leading to finding values of the variables more efficiently.

Linear Equations

Linear equations are fundamental as they involve expressions where variables are raised to the power of one. These equations form straight lines when graphed, offering a direct visual sense of their solutions.
In our problem, the equation \( 7x + 13 = A(3x + 5) + B(x + 1) \) is ultimately dissected into linear components by choosing strategic substitutions, converting the system into individual linear equations like \( 6 = 2A \), allowing for straightforward resolution.

• Each equation's solution leads to insights only seen through solving linear interactions.
• The beauty of linear equations lies in their simplicity and elegance, making them accessible and easy to solve.

Understanding linear equations helps foster comprehension of larger, more complex mathematical systems and offers a robust tool for dealing with various algebraic problems.