Problem 18 - Elementary Algebra

Question

Find the domain of the equations. Assume that the independent variable is the variable that appears in the expression to the right of the equal sign. $$ t=\frac{5}{s-6} $$

Step-by-Step Solution

Answer

Question: Determine the domain of the given equation: \(t = \frac{5}{s-6}\). Answer: The domain of the given equation is all real numbers, except \(s \neq 6\).

Step 1: Identify places where the denominator can be zero.

We have the equation: $$ t = \frac{5}{s-6} $$ The denominator will be zero when: $$ s-6 = 0 $$ This occurs when \(s=6\).

Step 2: Find the domain by excluding the values that make the denominator equal to zero.

Since the equation is undefined when \(s=6\), we need to exclude that value from the domain of the equation. Thus, the domain consists of all real numbers except \(s=6\): $$ \text{Domain: } s \in \mathbb{R} \text{ and } s \neq 6 $$

Key Concepts

Algebraic Expressions

Algebraic expressions include constants, variables, and arithmetic operations such as addition, subtraction, and multiplication. They form the building blocks of algebra and greatly help us express mathematical ideas. To understand this better, let's take our expression from the problem: \( t = \frac{5}{s - 6} \). Here, the term \(s - 6\) is part of the algebraic expression in the denominator.

• **Constants**: These are fixed values. In our expression, \(5\) is a constant.
• **Variables**: These are symbols that represent numbers. In our case, \(s\) is the variable, which can take on any real number except 6.
• **Operations**: These involve addition, subtraction, multiplication, or division. In the denominator \(s - 6\), subtraction is used.

Algebraic expressions can vary in complexity, but understanding each component helps in tackling problems like finding domains and simplifying expressions gracefully.

Rational Functions

Rational functions are fractions made up of two polynomial expressions. Specifically, you deal with a numerator and a denominator. The "rational" part is because it's like a fraction: rational numbers are ratios, and this is a ratio of two polynomials.In the exercise at hand, the function \( t = \frac{5}{s-6} \) is a rational function:

• The **numerator** is \(5\), a constant polynomial.
• The **denominator** is \(s-6\), a linear polynomial.

Rational functions have restrictions. Mainly, the denominator cannot be zero since division by zero is undefined. To find these restrictions, set the denominator equal to zero and solve for the variable. This tells you which values to exclude from the domain.In our example, the solution highlights that the denominator \(s-6\) becomes zero when \(s=6\), so \(s=6\) is not allowed in the domain. Being able to find these restrictions ensures that you create a valid set of inputs for the function.

Real Numbers

Real numbers include all the rational and irrational numbers. They form the most comprehensive set of numbers that we commonly encounter in mathematics, essential in describing measurable quantities. Some features of real numbers:

• **Rational numbers**: Can be expressed as fractions, including integers and finite or recurring decimals.
• **Irrational numbers**: Cannot be expressed as simple fractions, such as \(\sqrt{2}\) or \(\pi\).

In the realm of rational functions, the domain may extend across the set of real numbers, with specific exceptions where the expression is undefined.For the expression \( t = \frac{5}{s-6} \), the domain includes all real numbers except for \(s=6\). This means \(s\) can take on almost any "real" value--positive, negative, zero, or any decimal--except the single exclusion point calculated. Understanding the role of real numbers prepares you to handle diverse and complex problems involving varied numerical values.