Problem 43 - Elementary and Intermediate Algebra

Question

Solve each equation and check the result. If an equation has no solution, so indicate. $$ \frac{x+6}{x+4}+\frac{1}{x^{2}+x-12}=1 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{5}{2} \).

Step 1: Identify the Denominators

To solve the equation \( \frac{x+6}{x+4} + \frac{1}{x^{2}+x-12} = 1 \), first identify the denominators of each term. The denominators are \( x+4 \) and \( x^2 + x - 12 \).

Step 2: Factor the Quadratic Denominator

Factor the quadratic expression in the second fraction: \( x^2 + x - 12 \). It factors into \( (x+4)(x-3) \).

Step 3: Find a Common Denominator

The common denominator for the fractions is the product \( (x+4)(x-3) \) since it includes both the simple linear term \( x+4 \) and its presence in the factored quadratic.

Step 4: Rewrite Each Fraction

Express each fraction over the common denominator \( (x+4)(x-3) \): \( \frac{(x+6)(x-3)}{(x+4)(x-3)} + \frac{1}{(x+4)(x-3)} = \frac{(x+4)(x-3)}{(x+4)(x-3)} \).

Step 5: Combine the Fractions

Combine the fractions over the common denominator: \( \frac{(x+6)(x-3) + 1}{(x+4)(x-3)} = 1 \).

Step 6: Eliminate the Denominator

Since the denominators of both sides of the equation are the same and non-zero, multiply both sides by \( (x+4)(x-3) \) to eliminate the denominators: \((x+6)(x-3) + 1 = (x+4)(x-3) \).

Step 7: Simplify and Expand

Expand both sides: \( (x+6)(x-3) + 1 = (x^2 + x - 12) \). Expanding the left side gives \( x^2 - 3x + 6x - 18 + 1 = x^2 + 3x - 17 \).

Step 8: Set and Solve for x

Set the equations equal: \( x^2 + 3x - 17 = x^2 + x - 12 \). Simplify by cancelling \( x^2 \) from both sides: \( 3x - 17 = x - 12 \). Solve for \( x \) by subtracting \( x \) and adding 17 to both sides: \( 2x = 5 \), so \( x = \frac{5}{2} \).

Step 9: Check for Extraneous Solutions

Verify that \( x = \frac{5}{2} \) does not make any denominators zero: \( \frac{5}{2} + 4 eq 0 \), \( (\frac{5}{2})^2 + \frac{5}{2} - 12 eq 0 \). The solution is valid.

Key Concepts

Factoring Quadratics

Solving rational equations often involves expressions where a portion needs factoring. Factoring quadratics is crucial for this.
A quadratic expression is typically in the form \( ax^2 + bx + c \). To factor it, identify two numbers that multiply to \( ac \) and add up to \( b \).
In our exercise, we have \( x^2 + x - 12 \). Here, the task is to find two numbers that multiply to \(-12\) (the product of 1 and -12) and add to 1.

• The numbers \( +4 \) and \( -3 \) fit these criteria as \( +4 \times -3 = -12 \) and \( +4 + -3 = 1 \).
• Thus, the expression factors into \( (x+4)(x-3) \). Understanding this step ensures proper use of the quadratic in subsequent calculations.

When you can't easily identify the factors, try different combinations systematically or use the quadratic formula to help. It's a mathematical puzzle of sorts, and getting good at it takes practice. Factoring reveals simplified expressions, making them much easier to work with in equations.

Common Denominators

In rational equations, to combine or simplify fractions, find and use a common denominator.
This technique works like solving a puzzle, where you use numbers or expressions divisible by all the involved denominators.

• In our case, the denominators are \( x+4 \) and \( (x+4)(x-3) \).
• The common denominator must include each unique factor at least once. Here, it is \( (x+4)(x-3) \).

To rewrite each fraction with this common denominator, ensure that each fraction's numerator also accounts for the factor it was missing initially. Like adjusting a recipe to match a larger group, changing the numerators ensures both fractions are still equivalent to their original form. This step allows you to focus on comparing and solving the numerators, as the denominators now offer a uniform backdrop. It's an essential part for rational equations, easing further steps like combining fractions and ultimately simplifying the expression.

Extraneous Solutions

When solving equations, especially rational ones, extraneous solutions can occasionally appear.
These are results that solve your manipulated equation but don't satisfy the original one due to restrictions like division by zero.

• After solving for \( x \), it is essential to substitute back into the original denominators.
• Confirm that none become zero. In our case, check \( x+4 \) and \( x^2 + x - 12 = (x+4)(x-3) \).

If substituting leads a denominator to zero, then that solution isn't valid because dividing by zero is undefined in mathematics. For instance, if your solution made \( x+4 \) zero, that \( x \) was extraneous. Checking these values isn't just a step; it's a validation.
This ensures that the solutions make the original equation logical and meaningful.