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Magazine Chapter 7: Problem 18
$$ \text { For Problems 1-32, solve each equation. (Objective 1) } $$ $$ \frac{3 x-1}{x^{2}-9}+\frac{4}{x+3}=\frac{5}{x-3} $$
Short Answer
Expert verified
The solution is \(x = 14\).
Step by step solution
01
Identify Common Denominator
The denominators are \( x^2 - 9 \), \( x + 3 \), and \( x - 3 \). Recognize that \( x^2 - 9 \) can be factored as \((x+3)(x-3)\). The common denominator is \((x+3)(x-3)\).
02
Rewrite Each Fraction
Express each fraction with the common denominator \((x+3)(x-3)\). The first fraction is already under this denominator. The second fraction \(\frac{4}{x+3}\) becomes \(\frac{4(x-3)}{(x+3)(x-3)}\). The third fraction \(\frac{5}{x-3} \) becomes \(\frac{5(x+3)}{(x+3)(x-3)}\).
03
Combine Fractions
Combine all fractions into a single equation: \(\frac{3x-1 + 4(x-3)}{(x+3)(x-3)} = \frac{5(x+3)}{(x+3)(x-3)}\).
04
Simplify the Numerator
Simplify the expressions in the numerator: \(3x-1 + 4(x-3) = 3x-1 + 4x - 12 = 7x - 13\).
05
Solve the Equation
The equation is now \(\frac{7x - 13}{(x+3)(x-3)} = \frac{5(x+3)}{(x+3)(x-3)}\). Since the denominators are equal, equate the numerators: \(7x - 13 = 5(x+3)\).
06
Expand and Simplify
Expand the right side: \(5(x+3) = 5x + 15\). Simplify the equation by subtracting \(5x\) from both sides: \(7x - 5x - 13 = 15\) which gives \(2x - 13 = 15\).
07
Solve for \(x\)
Add 13 to both sides to isolate terms involving \(x\): \(2x = 28\). Divide both sides by 2 to solve for \(x\): \(x = 14\).
08
Check for Extraneous Solutions
Verify that \(x = 14\) satisfies the original equation and doesn't make any denominators zero. Inputting 14 into the denominators shows they are non-zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
When dealing with rational equations, factoring quadratics is often a crucial step. In our given problem, one of the denominators was a quadratic expression, specifically \(x^2 - 9\). To make solving the equation simpler, we factor this quadratic. Factoring means rewriting it as a product of simpler expressions.
For \(x^2 - 9\), we identify it as a difference of squares because it fits the form \(a^2 - b^2\), which factors into \((a + b)(a - b)\).
Therefore, \(x^2 - 9\) factors to \((x + 3)(x - 3)\).
For \(x^2 - 9\), we identify it as a difference of squares because it fits the form \(a^2 - b^2\), which factors into \((a + b)(a - b)\).
Therefore, \(x^2 - 9\) factors to \((x + 3)(x - 3)\).
- Understanding the difference of squares allows us to break down complex quadratics.
- This helps in finding a common denominator for rational equations.
Equation Solving Techniques
Solving rational equations involves several techniques that can make the process more methodical and successful. Here are some essential steps:
- Identify the Common Denominator: Unify the equation by finding a common denominator. In the exercise, this was \((x + 3)(x - 3)\).
- Rewriting Fractions: Adjust each fraction so that they all share this common denominator, which allows you to combine them into a single rational expression.
- Equate the Numerators: Once fractions have the same denominator, set the numerators equal to each other and solve like a regular algebraic equation.
Algebraic Fractions
Algebraic fractions, similar to numerical fractions, represent a division of expressions. They can initially appear daunting due to their variables and powers, but handle them just like number fractions. Here's how:
- Equivalent Algebraic Fractions: Express them with the same denominator to combine or compare them effectively.
- Operations with Algebraic Fractions: You can add, subtract, multiply or divide them using similar rules applied to numerical fractions.
- Reducing Algebraic Fractions: Simplify by factoring numerators and denominators, then canceling common factors, much like simplifying a numeric fraction.
Extraneous Solutions
When solving rational equations, checking for extraneous solutions is important to ensure that the solution truly satisfies the original equation. An extraneous solution is a solution that emerges during the process of solving but doesn't work when substituted back into the original equation.
Here's what to do:
Here's what to do:
- Identify which values make any denominators in the original equation zero. Any such values are not in the solution set.
- Substitute the found solution back into the original equation to verify it does not make any denominator zero.
- An extraneous solution is a result of multiplying through by a variable term, which can introduce invalid solutions.
