Problem 62
Question
Classify each equation as a contradiction, an identity, or a conditional equation. Give the solution set. Use a graph or table to support your answer. $$\frac{2 x-1}{3}=\frac{2 x+1}{3}$$
Step-by-Step Solution
Verified Answer
The equation is a contradiction with an empty solution set.
1Step 1: Identify and Simplify the Equation
We are given the equation \( \frac{2x - 1}{3} = \frac{2x + 1}{3} \). Since both sides of the equation have the same denominator, we can multiply both sides by 3 to eliminate the fractions, simplifying the equation to \( 2x - 1 = 2x + 1 \).
2Step 2: Isolate x
Next, we'll try to isolate \(x\) by subtracting \(2x\) from both sides of the equation: \(2x - 2x - 1 = 2x - 2x + 1\), which simplifies to \(-1 = 1\).
3Step 3: Classify the Equation
The equation \(-1 = 1\) is a contradiction because it is a false statement. This means there is no possible value of \(x\) that can satisfy the equation. Therefore, the solution set is empty.
Key Concepts
Understanding Contradictions in EquationsThe Notion of a Solution SetSimplifying Equations: The Basics
Understanding Contradictions in Equations
In mathematics, a contradiction occurs when an equation simplifies to an untrue statement. Unlike the usual arithmetic where equations are about finding values that make a sentence true, a contradiction is when the math tells us there’s no possible solution. For example, consider the equation \[-1 = 1.\] Clearly, this statement is false because negative one cannot equal positive one. When simplifying equations, as with our example equation given \[\frac{2x-1}{3} = \frac{2x+1}{3},\]you subtract terms and eventually end up with a false statement. This indicates that no number can satisfy the condition set by the equation, thus, we call this a contradiction. When faced with a contradiction, remember, the equation simply has no solution.
The Notion of a Solution Set
In mathematics, a solution set is the collection of all possible values that satisfy an equation. It tells us which numbers, if any, solve the equation. For equations that are contradictions, just like in our case of \[\frac{2x-1}{3} = \frac{2x+1}{3},\]after simplifying we get \[-1 = 1,\] suggesting no solution exists. This leads us to an empty solution set, denoted by \[\emptyset.\] In other words, there are no values of \(x\) that can make the equation true. In broader terms:
- If an equation is always true, regardless of the value of \(x\), the solution set contains every number.
- If specific numbers satisfy the equation, those numbers form the solution set.
- In contradictions, however, the solution set remains empty.
Simplifying Equations: The Basics
Simplifying equations is a fundamental step in solving mathematical problems. It's like cleaning up the clutter to see what's really happening. In essence, we perform operations to make the equation more straightforward or even solve it. Let's break it down:
- **Remove fractions**: Multiply both sides by the least common multiple to eliminate denominators.
- **Collect like terms**: Bring similar terms together, such as adding all \(x\) terms and constant terms separately to one side.
- **Isolate variables**: Get the variable \(x\), or any variable of interest, by itself on one side of the equation.
Other exercises in this chapter
Problem 61
Find \(f(x)\) at the indicated value of \(x\). $$f(x)=\sqrt{x^{3}+12} ; x=-2$$
View solution Problem 61
Find a decimal approximation of each root or power Round answers to the nearest thousandth. $$19^{1 / 2}$$
View solution Problem 62
Investment problems such as those in Exercises \(61-66\) can be solved by using a method similar to the one explained in Example \(2,\) along with the simple- i
View solution Problem 62
Find \(f(x)\) at the indicated value of \(x\). $$f(x)=\sqrt[3]{x^{2}-x+6} ; x=2$$
View solution