Problem 43 - College Algebra with Modeling and Visualization

Question

Solve each equation or inequality. (a) \(|2 x-3|=1\) (b) \(|2 x-3|<1\) (c) \(|2 x-3|>1\)

Step-by-Step Solution

Answer

(a) \(x = 1\) or \(x = 2\) (b) \(1 < x < 2\) (c) \(x < 1\) or \(x > 2\)

Step 1: Understand the Absolute Value Equation

When solving an equation of the form \(|a| = b\), where \(b\) is a non-negative number, it translates to two linear equations: \(a = b\) and \(a = -b\). For our equation, \(|2x - 3| = 1\), this implies we need to solve: \(2x - 3 = 1\) and \(2x - 3 = -1\).

Step 2: Solve for \(x\) in \(2x - 3 = 1\)

To isolate \(x\), add 3 to both sides of the equation to get \(2x = 4\). Then, divide both sides by 2 to find \(x = 2\).

Step 3: Solve for \(x\) in \(2x - 3 = -1\)

Add 3 to both sides of the equation, giving \(2x = 2\). Then, divide by 2 to find \(x = 1\).

Step 4: Solve the Absolute Value Inequality \(|2x - 3| < 1\)

The inequality \(|a| < b\) translates to \(-b < a < b\). So, \(|2x - 3| < 1\) becomes \(-1 < 2x - 3 < 1\).

Step 5: Solve the Compound Inequality \(-1 < 2x - 3 < 1\)

First, add 3 to all parts of the inequality: \(2 < 2x < 4\). Then, divide all parts by 2: \(1 < x < 2\).

Step 6: Solve the Absolute Value Inequality \(|2x - 3| > 1\)

For \(|a| > b\), the inequality splits into two: either \(a > b\) or \(a < -b\). So, \(|2x - 3| > 1\) implies \(2x - 3 > 1\) or \(2x - 3 < -1\).

Step 7: Solve \(2x - 3 > 1\)

Add 3 to both sides to obtain \(2x > 4\). Divide by 2 to find \(x > 2\).

Step 8: Solve \(2x - 3 < -1\)

Add 3 to both sides, resulting in \(2x < 2\). Divide by 2 to find \(x < 1\).

Step 9: Combine Solutions for \(|2x - 3| > 1\)

Combine the solutions: \(x > 2\) or \(x < 1\).

Key Concepts

Compound Inequalities

Understanding compound inequalities involves working with two inequalities at once. The solution encompasses a range where both conditions must be satisfied simultaneously, or it can refer to a situation where either one of two separate conditions must be satisfied.
This becomes particularly important when solving absolute value inequalities, such as \(|2x - 3| < 1\).
Here’s how it usually breaks down:

• For \(|a| < b\), it becomes \(-b < a < b\), implying a single range where both inequalities must be true simultaneously.
• For \(|a| > b\), it implies two separate cases: either \(a > b\) or \(a < -b\). Here, falling within either range satisfies the condition.
  

Thus, for \(|2x - 3| < 1\), we solved \(-1 < 2x - 3 < 1\) and found \(1 < x < 2\). This means that only values of \(x\) between 1 and 2 will make the original inequality true.

Linear Equations

Linear equations form the backbone of solving absolute value problems. These are equations of the first degree, which means the variables are not raised to any powers other than 1. In its simplest form, it looks like \(ax + b = 0\).
Understanding how to manipulate these equations is crucial. For instance, given an equation like \(2x - 3 = 1\), here's how it's handled:

• Add or subtract constants to isolate the term with the variable. Here, adding 3 to get \(2x = 4\).
• Divide by the coefficient of the variable to solve for \(x\), like dividing by 2 to find \(x = 2\).

Mastering these concepts allows students to efficiently solve equations that form the equality part of absolute value problems.

Solving Inequalities

The process of solving inequalities involves finding all possible solutions that make an inequality true when substituted into it.
It extends beyond solving linear equations as it deals with ranges or sets of numbers rather than just a single solution.

• For \(a < b\), find all values where \(a\) is less than \(b\). For instance, the inequality \(2x < 4\) translates to \(x < 2\).
• For \(a > b\), determine when \(a\) is greater than \(b\), like \(2x > 4\) meaning \(x > 2\).

Pay special attention when manipulating inequalities: When multiplying or dividing by a negative number, the inequality sign flips. For the compound inequalities stemming from absolute value inequalities, recognizing how to split them into manageable parts helps to find the complete solution set.