Problem 1 - Algebra 2 and Trigonometry

Question

Danielle said that there is no integer that makes the inequality \(|2 x+1|

Step-by-Step Solution

Answer

Danielle is correct; no integer satisfies \(|2x + 1| < x\).

Step 1: Understand the Inequality

The inequality given is \(|2x + 1| < x\). Our goal is to determine if there are any integers \(x\) that satisfy this inequality.

Step 2: Setting Up Conditions from Absolute Value

The expression \(|2x + 1|\) represents the absolute value, which means the expression can be split into two cases: 1. \(2x + 1 < x\) for the positive part, 2. \(-(2x + 1) < x\) for the negative part.

Step 3: Solving Case 1: \(2x + 1 < x\)

Subtract \(x\) from both sides: \(2x + 1 - x < 0\). This simplifies to \(x + 1 < 0\). Subtract 1 from both sides: \(x < -1\). So, any integer solution must be less than \(-1\).

Step 4: Solving Case 2: \(-(2x + 1) < x\)

This can be rewritten as \(-2x - 1 < x\). Add \(2x\) to both sides: \(-1 < 3x\). Divide every term by 3: \(-\frac{1}{3} < x\). So, \(x\) must be greater than \(-\frac{1}{3}\).

Step 5: Finding the Intersection of Solutions

The solutions from the two cases are: 1. \(x < -1\) 2. \(x > -\frac{1}{3}\). These conditions cannot be true at the same time for any integer because there is no integer both less than \(-1\) and greater than \(-\frac{1}{3}\).

Step 6: Conclusion and Verification with an Example

Since there is no intersection of valid integer solutions from both cases, no integer can satisfy the original inequality \(|2x + 1| < x\). Thus, Danielle's statement is correct.

Key Concepts

Integer Solutions

When we talk about integer solutions in mathematics, we refer specifically to whole numbers that satisfy a given equation or inequality. In the case of the inequality \(|2x + 1| < x\), we are interested in finding which, if any, of the integer values of \(x\) make this inequality true. To start, we consider the integers which are essentially numbers like -2, -1, 0, 1, 2, and so on.

When we solve inequalities involving absolute values, as in this exercise, we often break them down into two separate case scenarios (much like opening a locked box with two keys). Each scenario gives us its conditions that \(x\) must meet. If no integer meets the condition from both cases simultaneously, then there is no integer solution.

Thus, for our specific problem, Danielle is correct: there are no integers that make \(|2x + 1| < x\) true because each case condition produces sets of numbers that do not overlap with each other when it comes to integers.

Inequality Solving

Inequality solving involves finding the set of numbers that satisfy the inequality condition. Inequalities are similar to equations, but they show a relationship where values are greater or less than others, rather than equal.

For this task, understanding how to solve inequalities with absolute values is crucial, which can initially seem daunting. We first interpret the absolute value expression \(|2x + 1|\) as two separate inequalities. The expression \(2x + 1\) itself can either be positive or negative.

• When \(2x + 1\) is positive, it must already be less than \(x\) for it to hold true.
• When \(2x + 1\) is negative, we flip it to the positive equivalent (because the negative of a number whose absolute value we take is positive) and compare that against \(x\).

In our specific example, case one was solving \(2x + 1 < x\) to find \(x < -1\) as a condition. The second case, \(-(2x + 1) < x\), led to \(x > -\frac{1}{3}\).

These carefully constructed steps cover all possible scenarios that could arise from the original inequality. Even though it might not seem like it, solving inequalities is primarily about ensuring we respect and balance both mathematical possibilities.

Algebraic Cases Analysis

Algebraic cases analysis is an approach used to solve problems with multiple scenarios or situations defined by the same expression. It is particularly useful when facing absolute value expressions. This is because absolute values, by their nature, involve both their positive and negative counterparts.

Taking a scenario and treating it in different cases allows us to solve for all possible outcomes of the inequality. In the given problem, the two cases were:

• \(2x + 1 < x\) which arises when \(2x + 1\) is assumed to be inherently positive.
• \(-(2x + 1) < x\) when \(2x + 1\) is assumed to be negative.

Both of these scenarios unveil a section of the number line where potential solutions lie. The beauty and logic behind algebraic cases analysis are that it allows consideration of every possible situation that the inequality might present, helping to comprehensively reject impossible solutions.

In the end, we combine ranges or inequalities from each scenario to see if there are common integer solutions. If there aren't, as in Danielle's statement, it's mathematical proof that indeed no integer exists to satisfy all parts of the inequality simultaneously.