Problem 31 - College Algebra with Modeling and Visualization

Question

Solve the inequality symbolically. Express the solution set in set-builder or interval notation. $$ 3 \leq \frac{1}{2} x+\frac{3}{4} \leq 6 $$

Step-by-Step Solution

Answer

The solution set is \( \left[\frac{9}{2}, \frac{21}{2}\right] \).

Step 1: Break Down the Double Inequality

We begin by breaking down the double inequality into two separate inequalities: \(3 \leq \frac{1}{2}x + \frac{3}{4}\) and \(\frac{1}{2}x + \frac{3}{4} \leq 6\). We will solve each inequality separately.

Step 2: Solve the First Inequality

Solve the first inequality \(3 \leq \frac{1}{2}x + \frac{3}{4}\). First, subtract \(\frac{3}{4}\) from both sides:\[3 - \frac{3}{4} \leq \frac{1}{2}x\]Convert 3 to a fraction with a denominator of 4 to subtract:\[\frac{12}{4} - \frac{3}{4} = \frac{9}{4}\]So, \(\frac{9}{4} \leq \frac{1}{2}x\). Multiply both sides by 2 to solve for \(x\):\[\begin{aligned}2 \cdot \frac{9}{4} &\leq x, \\frac{18}{4} &\leq x, \\frac{9}{2} &\leq x.\end{aligned}\]

Step 3: Solve the Second Inequality

Solve the second inequality \(\frac{1}{2}x + \frac{3}{4} \leq 6\). First, subtract \(\frac{3}{4}\) from both sides:\[\frac{1}{2}x \leq 6 - \frac{3}{4}\]Convert 6 to a fraction with a denominator of 4 for subtraction:\[\frac{24}{4} - \frac{3}{4} = \frac{21}{4}\]Thus, \(\frac{1}{2}x \leq \frac{21}{4}\). Multiply both sides by 2 to solve for \(x\):\[\begin{aligned}2 \cdot \frac{1}{2}x &\leq 2 \cdot \frac{21}{4}, \x &\leq \frac{42}{4}, \x &\leq \frac{21}{2}.\end{aligned}\]

Step 4: Combine Solutions

Combine the solutions from both inequalities: \( \frac{9}{2} \leq x \leq \frac{21}{2} \). This represents the intersection of both inequality solutions.

Step 5: Express the Solution in Interval Notation

The solution set in interval notation is \[ \left[\frac{9}{2}, \frac{21}{2}\right] \]. This interval represents all values of \(x\) that satisfy both inequalities.

Key Concepts

Interval Notation

When expressing solution sets for inequalities, interval notation is a concise and visually intuitive method. It uses brackets to define the set of numbers within a certain range. Interval notation can show whether endpoints are included or excluded from the set.
For example:

• "\([a, b]\)" means all numbers between \(a\) and \(b\), including both \(a\) and \(b\).
• "\((a, b)\)" shows all numbers between \(a\) and \(b\), but does not include \(a\) or \(b\).
• "\([a, b)\)" includes \(a\) but not \(b\).
• "\((a, b]\)" includes \(b\) but not \(a\).

In our solution, \(\left[\frac{9}{2}, \frac{21}{2}\right]\) tells us that all numbers \(x\) from \(\frac{9}{2}\) to \(\frac{21}{2}\) are solutions to the inequality, and it includes both endpoints.

Algebraic Inequalities

Algebraic inequalities involve expressions with variables that use inequality signs instead of equal signs. These signs show a relationship that does not require equality, such as "greater than" or "less than."
The basic inequality symbols include:

• "\(<\)" (less than)
• "\(>\)" (greater than)
• "\(\leq\)" (less than or equal to)
• "\(\geq\)" (greater than or equal to)

A double inequality, like in our problem, expresses two inequality statements involving the same variable. The goal is to solve these separately and then combine the results. By breaking complex inequalities into manageable parts, solving them becomes easier. You just need to remember to adjust both parts of the inequality in the same manner, ensuring any operations conducted maintain the equality.

Solution Sets

Solution sets describe all possible values that satisfy a given inequality. Once you've solved the inequality, you express the set of solutions in a clear format, often using either set-builder notation or interval notation.
When using set-builder notation, you describe the set by indicating the property that elements follow. For example, "\( \{x | \text{condition} \} \)" tells us that \(x\) must meet the condition inside the braces.
Interval notation, which we used here, is handy for visually conveying the continuous range of possible values. In the example \( \left[\frac{9}{2}, \frac{21}{2}\right] \), the solution set includes every number between \( \frac{9}{2} \) and \( \frac{21}{2} \), inclusive of the bounds. This compact form makes it straightforward to understand what values of \(x\) the inequality encompasses.