Problem 33 - College Algebra and Calculus: An Applied Approach

Question

Solve the inequality. Then graph the solution set on the real number line. \(-10 x<40\)

Step-by-Step Solution

Answer

The solution of the inequality is \((- \infty, -\frac{3}{2})\), \((-5, -1)\), \((-1, +\infty)\)

Step 1: Subtract One Fraction from the Other

To simplify, move the fractions on one side, so the inequality will look like this: \[\frac{4}{x+5} - \frac{1}{2x+3} > 0\]

Step 2: Find common denominator

Using \( (x + 5) \) and \( (2x + 3) \) as denominators, multiply top and bottom of first fraction by \( 2x + 3 \) and the second fraction by \( x + 5 \) to find common denominator:\[\frac{4(2x+3)}{(x + 5)(2x + 3)} - \frac{(x + 5)}{(x + 5)(2x + 3)} > 0\]

Step 3: Simplify the Numerators

Simplify the numerators, resulting in:\[\frac{8x+12-x-5}{(x + 5)(2x + 3)} > 0\]which simplifies further to\[\frac{7x+7}{(x + 5)(2x + 3)} > 0\]Removing the common factor of 7 in the numerator:\[\frac{x+1}{(x + 5)(2x + 3)} > 0\]

Step 4: Identify the Critical Points

Set each part of the factorized inequality to zero, identifying the critical points:\[x+1 = 0, \ x = -1\]\[x+5 = 0, \ x = -5\]\[2x + 3 = 0, \ x = -\frac{3}{2}\]

Step 5: Find solution regions

The solution of the inequality will fall within intervals determined by the critical points. Plug in sample values from these intervals into the inequality and check the sign. We end up with the following regions: \((- \infty, -\frac{3}{2})\), \((-5, -1)\), \((-1, +\infty)\)

Step 6: Graph the Solution

On a number line, graph the solution set, which would include the intervals (-infinity, -3/2), (-5, -1), and (-1, infinity). Note, however, that the inequality does not include the critical points itself, so these points will be open circles on the number line.

Key Concepts

Real number line

The real number line is a visual way to represent all real numbers in a continuous line. On this line, every point corresponds to a real number, and it is often used to graphically display the solution sets of equations and inequalities.

In solving inequalities, like the one given in the exercise, the real number line helps us portray complex expressions in an accessible form. The points on the line act as critical markers indicating where inequalities change from true to false or vice versa. To show a solution set, intervals are marked on the line between these critical points.

For example, if you need to indicate that an inequality is true for all numbers greater than zero, you can shade the line right from zero, often using an open circle to show that zero itself is not included. This approach simplifies complex mathematical concepts into something more visually tangible for students.

Critical points

Critical points are values of the variable that make the numerator or denominator equal to zero. In inequalities, these points are essential because they determine the boundaries where the inequality changes. Let's break it down further:

• Set the numerator equal to zero to find when the expression itself is zero.
• Set the denominator equal to zero to find values that make the expression undefined.

In the given inequality, \[\frac{x+1}{(x + 5)(2x + 3)} > 0\]the critical points are found by solving:- \(x+1 = 0\) giving \(x = -1\) - \(x+5 = 0\) giving \(x = -5\)- \(2x+3 = 0\) giving \(x = -\frac{3}{2}\)
These points divide the number line into intervals that we test to determine the sign of the inequality in each interval. Critical points help us understand where significant changes occur in the behavior of the inequality.

Fraction manipulation

In the process of solving inequalities involving fractions, manipulating the fractions correctly accounts for a significant part of the solution. Here, each fraction must be adjusted to have a common denominator so that the inequality can be simplified effectively.

To combine \[\frac{4}{x+5} - \frac{1}{2x+3} > 0\]a common denominator of \((x+5)(2x+3)\) is used. Multiplying top and bottom of each fraction separately by appropriate terms ensures that the fractions have a common base.

Like this:

• Multiplied \(4/(x+5)\) by \((2x+3)/(2x+3)\)
• Multiplied \(1/(2x+3)\) by \((x+5)/(x+5)\)

This creates a single simplified differential fraction that you can analyze by zeroing in on numerators and factoring. Mastery of fraction manipulation is a key skill in algebra that allows the simplification of complex expressions, leading to a deeper understanding of both equations and inequalities.

Solution set

A solution set is a range of values that satisfy the given inequality. Once critical points are identified, testing these values help locate regions in the inequality that hold true.

In this exercise, once we rearrange the inequality and identify critical points (\(-5\), \(-\frac{3}{2}\), and \(-1\)),we determine which intervals satisfy the inequality by testing values and checking signs.

• For \((-\infty, -\frac{3}{2})\), pick \(x = -2\)
• For \((-5, -1)\), pick \(x = -3\)
• For \((-1, +\infty)\), pick \(x = 0\)

Each test will reveal whether the inequality holds true in that interval. The solution set excludes critical points where the fraction is undefined or changes sign; hence, marked as open circles on the number line. This visual and analytical process gives students a complete understanding of which values of \(x\) satisfy the inequality.