Problem 39 - College Algebra Essentials

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. \(\frac{1}{x-1}+5=\frac{11}{x-1}\)

Step-by-Step Solution

Answer

The restriction on \(x\) is that it cannot be equal to \(1\). The solution to the equation is \(x = 3\).

Step 1: Identify the Restriction

In the equation \(\frac{1}{x-1}+5=\frac{11}{x-1}\), the denominators are \(x - 1\). A denominator cannot be zero, so we must find the value that would make this happen. We solve for \(x\) in \(x - 1 = 0\). Adding 1 to both sides, we find that \(x = 1\).

Step 2: Solving the Rational Equation

Now, we proceed with solving the equation As long as \(x\neq1\) (our restriction). Rewrite \(\frac{1}{x-1}+5=\frac{11}{x-1}\) as \(\frac{1}{x-1} + \frac{5(x-1)}{x-1} = \frac{11}{x-1}\), which gives \(\frac{1+5x-5}{x-1} = \frac{11}{x-1}\). Reducing the left side to \(\frac{5x-4}{x-1} = \frac{11}{x-1}\). In a situation when we have equality with the same denominators, it means the numerators are also equal. Hence, \(5x-4=11\). Solving for \(x\), we add 4 to both sides, which gives us \(5x = 15\). Further dividing by 5, we find that \(x = 3\).

Key Concepts

Variable Restrictions

In a rational equation, restricting the variable is critical. When there is a variable in the denominator, we need to ensure it does not make the denominator zero. This is because division by zero is undefined in mathematics.
To find these restrictions, we set the denominator equal to zero and solve for the variable.
From our exercise with the equation \( \frac{1}{x-1}+5=\frac{11}{x-1} \), the denominator for both terms is \( x-1 \). We set \( x-1=0 \), and find \( x=1 \). This means \( x \) cannot be 1, as it would make the denominator zero, causing the equation to be undefined.
Identifying these restrictions first ensures you're working with a valid mathematical statement as you proceed to solve the equation.

Solving Equations

Solving rational equations involves finding the variable that satisfies the equation considering any restrictions.
Once you've determined variable restrictions, you can manipulate the equation to isolate the variable. In our exercise, first express all terms with a common denominator,\( x-1 \), which simplifies combining like terms.
For the equation \( \frac{1}{x-1} + 5 = \frac{11}{x-1} \), write it as \( \frac{1 + 5(x-1)}{x-1} = \frac{11}{x-1} \), which simplifies to \( \frac{5x - 4}{x-1} = \frac{11}{x-1} \).
The key is to simplify the expression and then equate numerators since the denominators are equal.

Denominators

Denominators play a vital role in rational equations. They determine the expression over which we need to maintain certain conditions such as not being equal to zero.
In the example \( \frac{1}{x-1}+5=\frac{11}{x-1} \), both sides of the equation have a common denominator \( x-1 \).
This allows us to manage the terms effectively and equates the numerators directly once the setups are equal. It provides a pathway for simplifying and eventually solving the equation. Understanding denominators are crucial for rewriting terms and making the equation solvable. Remember, we focus on maintaining the equation within its valid domain.

Equation Equality

Equation equality in rational expressions means both sides of the equation have the same value once simplified.
When the denominators are the same, equation equality allows us to focus solely on the numerators. In our specific case, upon simplifying, the equation becomes \( \frac{5x-4}{x-1} = \frac{11}{x-1} \).
Hence, we relate the equality of equations by equating \( 5x - 4 \) to \( 11 \). Solving this gives us \( 5x=15 \), leading to \( x=3 \). This step calculates the variable solution that holds the equation true, ensuring compliance with any prior restrictions discussed.