Problem 49 - Algebra and Trigonometry

Question

Contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2 x-8}$$

Step-by-Step Solution

Answer

The restrictions of the variable are \(x \neq 4\) and \(x \neq -2\). However, upon solving the equation, the value of \(x\) comes out to 4 which is one of the restrictions. Therefore, this equation has no solutions.

Step 1: Identify Restrictions

First, set each denominator equal to zero and solve for \(x\). This gives us the restrictions on the variable. \[x - 4 = 0 \Rightarrow x = 4\] \[x + 2 = 0 \Rightarrow x = -2\] \[x^{2} - 2x - 8 = 0 \Rightarrow x = 4, -2\] So, \(x \neq 4\) and \(x \neq -2\]. These are the values which cannot be taken by \(x\) as they will make at least one of the denominators zero, causing undefined values in the equation.

Step 2: Solve the Equation

In the next step, change the original equation, \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2 x-8}\), to a simpler equation without denominators by multiplying both sides by the common denominator, \(x^{2} - 2x - 8\). This generates an equivalent equation without any denominators: \(x + 2 - 5(x - 4) = 6\). Solve this equation: - Distribute the 5 on the left side to get: \(x + 2 - 5x + 20 = 6\). - Combine like terms to simplify and get the equation in the form \(ax + b = c\): \(-4x+22=6\). - Subtract 22 from both sides: \(-4x = -16\). - Divide both sides by -4 to solve for \(x\): \(x = 4\).

Key Concepts

Restrictions on Variables

In rational equations, restrictions on variables are crucial because they identify values that would make the denominator zero. Having a zero in the denominator leads to an undefined expression, which is mathematically invalid. To determine these restrictions, you set each denominator equal to zero and solve for the variable.

• For example, in the equation \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2x-8}\), you need to make sure that the denominators, \(x-4\), \(x+2\), and \(x^{2}-2x-8\), do not equal zero.
• Setting \(x-4=0\), we find \(x=4\).
• Solving \(x+2=0\), gives \(x=-2\).


• Finally, the quadratic \(x^{2}-2x-8=0\) also results in \(x=4, -2\).

Thus, the values \(x=4\) and \(x=-2\) are restrictions and should be avoided in solutions, as they would make the equation undefined.

Solving Equations

The main goal in solving rational equations is to isolate the variable, typically through manipulation that removes denominators. To tackle the equation \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2x-8}\), you simplify by finding an equivalent equation without denominators.
A common method is to multiply every term by the least common denominator (LCD). This eliminates fractions, allowing easier manipulation.

• For this equation, the LCD is \(x^{2}-2x-8\).
• Multiply every term by this LCD to clear denominators.

The resulting equation, \(x + 2 - 5(x - 4) = 6\), contains no fractions and is simpler. Once transformed, solve the equation as you would any linear equation by distributing, combining like terms, isolating the variable, and solving for \(x\).

Common Denominator

In rational equations, finding a common denominator is fundamental for clearing fractions. A common denominator ensures that you can perform operations across fractions without compromising the equation. It’s effectively the smallest expression that each denominator divides into without a remainder.
In the rational equation \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2x-8}\), the common denominator is \(x^{2}-2x-8\). Finding this ensures you properly eliminate denominators so each term can be considered equally.

• Multiply each fraction in the equation by this common denominator.
• This creates an equivalent equation without fractions, simplifying the solving process.

Working through this step lets us move towards finding a solution without the complications that fractions introduce.

Undefined Values

Undefined values in rational equations occur when a denominator equals zero. Mathematics doesn’t allow division by zero, labeling any calculations resulting from this as undefined.
When solving \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2x-8}\), it’s necessary first to recognize these undefined values. These cause restrictions such that \(x\) cannot equal 4 or -2, as derived from solving the denominators for zero. If inadvertedly we include these values in our solution set, the equation will not hold since division by zero is nonsensical.
Always verify your solution doesn't include these values by checking your final answers against the restrictions identified earlier. This ensures all solutions are valid and the equation remains defined throughout.