Problem 94 - A Graphical Approach to College Algebra

Question

Solve each inequality analytically. Write the solution set in interval notation. Support your answer graphically. $$\frac{2 x+3}{5}-\frac{3 x-1}{2}<\frac{4 x+7}{2}$$

Step-by-Step Solution

Answer

The solution set is \( \left(-\frac{24}{31}, \infty \right) \).

Step 1: Clear the Fractions

To eliminate the fractions, first determine the least common denominator (LCD) of the fractions involved. The fractions are over 5 and 2, thus the LCD is 10. Multiply through by 10 to clear the fractions.\[ 10 \left( \frac{2x+3}{5} \right) - 10 \left( \frac{3x-1}{2} \right) < 10 \left( \frac{4x+7}{2} \right) \] Which gives:\[ 2 \times (2x+3) - 5 \times (3x-1) < 5 \times (4x+7) \]

Step 2: Expand the Terms

Distribute the multiplication across the terms within the brackets.\[ 4x + 6 - 15x + 5 < 20x + 35 \]

Step 3: Combine Like Terms

Combine the terms to simplify the inequality:\[ (4x - 15x) + (6 + 5) < 20x + 35 \] which simplifies to:\[ -11x + 11 < 20x + 35 \]

Step 4: Transpose the Terms

Move all terms involving \(x\) to one side and constant terms to the other side.\[ -11x - 20x < 35 - 11 \] Simplifying gives:\[ -31x < 24 \]

Step 5: Solve for x

Divide each side by -31 to isolate \(x\), recalling that dividing by a negative number reverses the inequality sign.\[ x > \frac{24}{-31} \] or \[ x > -\frac{24}{31} \]

Step 6: Express in Interval Notation

Now express the solution for \(x\) in interval notation. Since \(x\) is greater than \(-\frac{24}{31}\), the interval notation is:\[ \left( -\frac{24}{31}, \infty \right) \]

Step 7: Graph the Solution

Graph the solution on a number line. The graph should have an open circle at \(-\frac{24}{31}\) and shade to the right, indicating that \(x\) can be any value greater than \(-\frac{24}{31}\).

Key Concepts

Interval Notation

Interval notation is a way of representing a set of numbers along a number line. It simplifies the expression of a range, making the solution more readable. When writing in interval notation, you use either round brackets, \((..., ... )\), or square brackets, \([..., ... ]\). In this system:

• Round brackets denote that the endpoint is not included in the set, known as an open interval. For instance, \((a, b)\) indicates all numbers greater than \(a\) and less than \(b\).
• Square brackets show that the endpoint is included, referred to as a closed interval. As an example, \([c, d]\) includes all numbers from \(c\) to \(d\) including \(c\) and \(d\).

In our inequality solution, the final result was expressed as \(( -\frac{24}{31}, \infty )\). This tells us that the solutions include all numbers greater than \(-\frac{24}{31}\) but not \(-\frac{24}{31}\) itself. The infinity symbol, \(\infty\), always uses a round bracket because it represents an endless range.

Least Common Denominator

To solve inequalities involving fractions, one effective step is to first clear the fractions using the least common denominator (LCD). The LCD is the smallest multiple that is common among the denominators in the given expressions. For the fractions in our exercise, the denominators were 5 and 2. The least common denominator for these is 10.

Why do we multiply by the LCD? Multiplying every term by the LCD standardizes the form of the inequality, eliminating fractions and thus simplifying the expressions to handle easier. In mathematical terms, if you have:

• \(\frac{a}{5} - \frac{b}{2}\), you multiply each fraction by 10 to clear the denominators.
• This results in \(10 \times \frac{a}{5} - 10 \times \frac{b}{2}\), which simplifies to \(2a - 5b\).

By resolving the fractions, the focus shifts to straightforward algebraic manipulation, making it easier to combine terms and solve for the variable.

Combining Like Terms

Combining like terms is an essential skill in solving inequalities as it leads to simpler expressions. Like terms are those terms in an expression that have the same variable to the same power. Let's see this in practice:

• Consider the inequality we have after clearing fractions: \(4x + 6 - 15x + 5 < 20x + 35\).
• The like terms involving \(x\) are \(4x\) and \(-15x\). These can be combined to give \(-11x\).
• The constant terms, \(6\) and \(5\), can be combined to yield \(11\).
• This reduces our inequality to the simpler form \(-11x + 11 < 20x + 35\).

Combining terms reduces complexity and prepares the inequality for the next steps such as isolating the variable, which aids in solving the inequality efficiently.

Graphical Solutions

Graphical solutions provide a visual representation of inequality solutions on a number line, making it easier to interpret and verify them. Using a graph helps to understand the range of solutions without complex calculations.

How do we do this? After solving the analytical part of the inequality, where we found \(x > -\frac{24}{31}\), we translate it onto a number line.

• You place an open circle at \(-\frac{24}{31}\) to indicate this point is not included in the solution set.
• You then shade the number line to the right of this point, showing that all numbers greater than \(-\frac{24}{31}\) are part of the solution.

Graphical solutions are particularly useful for visually illustrating the inequality's behavior and confirming the analytical solution's correctness. It's a good practice to draw this visual cue to ensure you've interpreted the results correctly.