Problem 19 - Algebra A Combined Function

Question

Solve each equation and check each proposed solution. See Examples 4 through 6. $$ \frac{1}{x+3}+\frac{6}{x^{2}-9}=1 $$

Step-by-Step Solution

Answer

The solution is \(x=4\). \(x=-3\) is invalid because it causes division by zero.

Step 1: Identify Common Denominator and Simplify

First, observe the denominators: \(x+3\) and \(x^2-9\). Notice that \(x^2-9\) is a difference of squares and can be written as \((x+3)(x-3)\). This reveals a common denominator: \((x+3)(x-3)\). Rewrite the equation in terms of this common denominator:\[\frac{1}{x+3} + \frac{6}{x^2-9} = 1\]This can be expressed as:\[\frac{(x-3) + 6}{(x-3)(x+3)} = 1\]

Step 2: Clear the Fractions

Multiply every term by the denominator \((x+3)(x-3)\) to eliminate the fraction:\[(x-3) + 6 = (x+3)(x-3)\]Simplify the equation to:\[x + 3 = x^2 - 9\]

Step 3: Rearrange to Form a Quadratic Equation

Bring all terms to one side to form a standard quadratic equation:\[0 = x^2 - x - 12\]Rearrange it as:\[x^2 - x - 12 = 0\]

Step 4: Solve Quadratic Equation Using Factoring

Factor the quadratic equation \(x^2 - x - 12 = 0\):\[(x-4)(x+3) = 0\]Set each factor equal to zero:1. \(x-4=0\): yields \(x=4\)2. \(x+3=0\): yields \(x=-3\)

Step 5: Check Solutions in Original Equation

We need to check each solution to ensure they satisfy the original equation \(\frac{1}{x+3} + \frac{6}{x^2-9} = 1\):1. For \(x=4\): - Plug \(x=4\) into the original equation, we get: \[ \frac{1}{4+3} + \frac{6}{16-9} = 1\to \frac{1}{7} + \frac{6}{7} = 1\to \frac{7}{7}=1 \] The equation is satisfied.2. For \(x=-3\): - Plug \(x=-3\) into the equation: \[ \frac{1}{-3+3} + \frac{6}{(-3)^2-9} = 1 \] This leads to division by zero in both terms, which is undefined. Therefore, \(x = -3\) is not a valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are a type of algebraic equation where the highest exponent of the variable is 2, typically taking the form \(ax^2 + bx + c = 0\). These equations appear frequently in algebra, and solving them is a fundamental skill in mathematics.
To solve a quadratic equation, we can use several methods, such as factoring, completing the square, or using the quadratic formula.
- **Factoring:** This involves writing the equation as a product of two binomials and setting each to zero.- **Quadratic Formula:** This method is applicable to all quadratic equations and is derived from the general form.- **Completing the Square:** This method rewrites the equation to make it a perfect square trinomial.In our exercise, we used factoring to solve the quadratic equation formed after clearing the fractions and rearranging terms. The key is to make sure the equation equals zero before attempting to factor.

Factoring

Factoring is the process of breaking down an expression into its simpler factors, which when multiplied together give the original expression. In algebra, especially when dealing with quadratic equations, factoring is often the quickest method for finding solutions.
For quadratics of the form \(x^2 - x - 12\), we look for two numbers that multiply to the constant term (-12) and add up to the linear term coefficient (-1). In this example:- The solution starts by identifying \(-3\) and \(+4\) as factors.- Therefore, the quadratic can be factored as \((x-4)(x+3) = 0\).Setting each factor to zero gives possible solutions for \(x\).
Factoring transforms the equation into a more manageable form, revealing the roots of the equation directly.

Common Denominator

Finding a common denominator is crucial when working with equations involving fractions. The common denominator allows you to combine and simplify fractions, making the equations easier to solve.
In the original equation, we had two denominators: \(x+3\) and \(x^2-9\). Recognizing that \(x^2-9\) can be factored into \((x+3)(x-3)\) helps us find a common ground for addition:

• Every term is rewritten with a denominator of \((x+3)(x-3)\).

This reveals a simplified and unified equation, enabling the removal of fractions by multiplying through by the common denominator. This step transforms the rational equation into a quadratic one, paving the way for standard solving techniques like factoring.

Solutions Checking

Solutions checking is an essential final step in solving equations. It ensures that the solutions satisfy the original equation, avoiding errors related to domain restrictions or arithmetic mistakes.
In our exercise, each proposed solution was plugged back into the original equation:

• For \(x=4\), the substitution verified that it satisfies the equation. The terms \(\frac{1}{4+3}\) and \(\frac{6}{16-9}\) simplified to 1.
• For \(x=-3\), direct substitution revealed a division by zero scenario in the original fractions, marking it as an invalid solution due to an undefined expression.

Checking solutions is crucial because solving algebraic fractions often involves potential restrictions, such as undefined values at certain inputs. This process confirms which solutions are truly valid.