Problem 43 - A Graphical Approach to College Algebra

Question

Solve each group of equations and inequalities analytically. (a) \(|2 x+1|+3=5\) (b) \(|2 x+1|+3 \leq 5\) (c) \(|2 x+1|+3 \geq 5\)

Step-by-Step Solution

Answer

(a) \(x = \frac{1}{2}, -\frac{3}{2}\); (b) \([-\frac{3}{2}, \frac{1}{2}]\); (c) \(x \geq \frac{1}{2}\) or \(x \leq -\frac{3}{2}\).

Step 1: Simplify the Equation

For equation (a), start with \(|2x+1|+3=5\). Subtract 3 from both sides to get \(|2x+1| = 2\).

Step 2: Solve the Absolute Value Equation

For \(|2x+1| = 2\), there are two cases: 1. \(2x+1 = 2\) which implies \(2x = 1\) and \(x = \frac{1}{2}\).2. \(2x+1 = -2\) which implies \(2x = -3\) and \(x = -\frac{3}{2}\).Thus, the solutions for (a) are \(x = \frac{1}{2}\) and \(x = -\frac{3}{2}\).

Step 3: Simplify the Inequality

For inequality (b), \(|2x+1|+3\leq5\). Subtract 3 from both sides to get \(|2x+1| \leq 2\).

Step 4: Solve the Absolute Value Inequality

For \(|2x+1| \leq 2\), there are two cases:1. \(2x+1 \leq 2\) which implies \(2x \leq 1\) and \(x \leq \frac{1}{2}\).2. \(2x+1 \geq -2\) which implies \(2x \geq -3\) and \(x \geq -\frac{3}{2}\).The solution for (b) is \(-\frac{3}{2} \leq x \leq \frac{1}{2}\).

Step 5: Simplify the Reverse Inequality

For inequality (c), \(|2x+1|+3\geq5\). Subtract 3 from both sides to get \(|2x+1| \geq 2\).

Step 6: Solve the Reverse Absolute Value Inequality

For \(|2x+1| \geq 2\), there are two cases:1. \(2x+1 \geq 2\) which implies \(2x \geq 1\) and \(x \geq \frac{1}{2}\).2. \(2x+1 \leq -2\) which implies \(2x \leq -3\) and \(x \leq -\frac{3}{2}\).The solution for (c) is \(x \geq \frac{1}{2}\) or \(x \leq -\frac{3}{2}\).

Key Concepts

Understanding Inequalities

Inequalities often involve expressions that are not equal and are represented with signs like \(>\), \(<\), \(\geq\), and \(\leq\). Handling inequalities is crucial because it allows us to understand or define ranges of solutions or conditions.
Two important types of inequalities you'll encounter are:

• Linear inequalities: They involve simple expressions without powers, like \(2x + 1 \leq 3\).
• Absolute value inequalities: Encompassing expressions such as \(|2x+1| \leq 5\), absolute value inequalities can be broken down into two scenarios to find a solution.

The absolute value \(|a|\) represents the distance from zero, making it always non-negative.
When solving \(|ax + b| \leq c\), you create two simple inequalities: \(ax + b \leq c\) and \(ax + b \geq -c\).
Similarly, for \(|ax + b| \geq c\), you solve for \(ax + b \geq c\) or \(ax + b \leq -c\). This approach helps us find the entirety of the solution set in practical terms.

Analytical Solving of Equations

Analytical solving involves a series of steps to distill an equation to its simplest form before finding solutions. It's a methodical approach that emphasizes understanding the transformations applied.
For the equation \(|2x+1|+3=5\), start by isolating the absolute value on one side.
Subtracting 3 from both sides gives \(|2x+1|=2\).
From here, break the absolute value equation into two separate linear equations:

• First, solve \(2x+1=2\) which gives \(x=\frac{1}{2}\).
• Second, solve \(2x+1=-2\) leading to \(x=-\frac{3}{2}\).

This technique of separating into cases helps ensure that both possible values that satisfy the equation are accurately captured.
Understanding that absolute value impacts how we interpret expressions under different conditions is the core of analytical solving.

Algebraic Solutions and their Simplification

Algebraic solutions involve using algebraic rules and identities to solve equations and inequalities efficiently. The process starts with simplifying the expression to an easily solvable form.
For instance, take \(|2x+1|+3 \leq 5\). First, eliminate the constant by subtracting 3 to convert it to \(|2x+1| \leq 2\).
Next, split this absolute value inequality into two algebraic expressions:

• \(2x+1 \leq 2\)
• \(2x+1 \geq -2\)

Solving these algebraic inequalities gives a range of solutions, \(-\frac{3}{2} \leq x \leq \frac{1}{2}\).
The spread of solutions shows that \(x\) can take any value within a specific domain, clearly indicating how inequalities can define solution intervals.
The reverse case of \(|2x+1| \geq 2\) can be split into expressions for different sections of the number line:

• \(2x+1 \geq 2\), resulting in \(x \geq \frac{1}{2}\)
• \(2x+1 \leq -2\), which gives us \(x \leq -\frac{3}{2}\)

Algebraic solutions provide a methodical path to capture all potential solutions for equations and inequalities efficiently.