Problem 64 - Intermediate Algebra

Question

Solve each compound inequality, if possible. Graph the solution set (if one exists) and write it using interval notation. \(5(x+1) \leq 4(x+3)\) and \(x+12<-3\)

Step-by-Step Solution

Answer

No solution; the intervals do not overlap.

Step 1: Distribute Terms in the First Inequality

The first inequality is \(5(x+1) \leq 4(x+3)\). Distribute the numbers outside the parentheses to simplify the expression. This gives us\[5x + 5 \leq 4x + 12.\]

Step 2: Solve the First Inequality

To isolate \(x\), subtract \(4x\) from both sides, which gives us \(x + 5 \leq 12\). Next, subtract \(5\) from both sides to solve for \(x\). This results in \[x \leq 7.\]

Step 3: Solve the Second Inequality

The second inequality is \(x + 12 < -3\). Subtract \(12\) from both sides to solve for \(x\). This leads to\[x < -15.\]

Step 4: Interpret the Compound Inequality

Now, consider the solutions \(x \leq 7\) from the first inequality and \(x < -15\) from the second inequality. Since both conditions must be true simultaneously, we need to find \(x\) values that satisfy both. However, \(x < -15\) is more restrictive and does not overlap with \(x \leq 7\).

Step 5: Represent Solution Using Interval Notation

The inequalities do not intersect because \(x < -15\) signifies values less than \(-15\), which are outside the range of \(x \leq 7\). Therefore, there is no solution that satisfies both conditions, meaning there is no interval that represents this compound inequality.

Key Concepts

Inequality Solving

When it comes to solving inequalities, the process is similar to solving equations, with a few important rules to remember. An inequality, like an equation, involves finding the values of a variable that make the inequality true. It is crucial to note the direction of the inequality sign when performing operations. For most operations, like addition, subtraction, multiplication, or division, the inequality sign stays the same. However, if you multiply or divide both sides by a negative number, the inequality sign flips.

In this exercise, we were given two inequalities:

• \(5(x+1) \leq 4(x+3)\)
• \(x+12<-3\)

Let's break them down one by one:

1. For the first inequality, start by distributing and simplifying to get \(5x + 5 \leq 4x + 12\). By subtracting \(4x\) from both sides, we isolated the variable which gave us \(x + 5 \leq 12\). Then, subtract \(5\) from both sides to solve for \(x\), leading to \(x \leq 7\).

2. For the second inequality, solve \(x + 12 < -3\) by subtracting \(12\), resulting in \(x < -15\).

Now, consider the conditions must hold for a true result simultaneously, but \(x < -15\) doesn't overlap with \(x \leq 7\). Hence, no solution satisfies both.

Interval Notation

Interval notation is an efficient way to write the solution sets of inequalities. It uses parentheses and brackets to describe a range of numbers, indicating the set of values that satisfy an inequality.

For example:

• \((a, b)\) represents all numbers between \(a\) and \(b\), not including \(a\) and \(b\).
• \([a, b]\) includes \(a\) and \(b\).
• An open interval, \((a, b)\), indicates elements greater than \(a\) and less than \(b\).
• A closed interval, \([a, b]\), includes end values as well.

In the case of compound inequalities, you often find the intersection (or overlap) of two intervals. However, in the given exercise, there's no overlap between the intervals defined by \(x \leq 7\) and \(x < -15\). Therefore, there is no solution, which we express as an empty set in interval notation:

• \(\emptyset\) or simply say there is no interval.

Graphing Inequalities

Graphing inequalities helps visualize solution sets on a number line, making it easier to see potential overlaps or intersections. Here’s how to approach it:

1. **Draw a number line:** Start by sketching a horizontal line and marking relevant points representing the boundaries or solutions of your inequalities.

• For the inequality \(x \leq 7\), place a point or line at \(7\) with a closed circle to show it includes \(7\) and shade to the left, indicating numbers less than or equal to \(7\).

• For \(x < -15\), mark \(-15\) with an open circle since it doesn't include \(-15\) and shade to the left for numbers less than \(-15\).

2. **Examine for overlap:** On the number line, observe both shaded regions. The absence of overlapping areas indicates no numbers satisfy both conditions simultaneously.

In this problem, you’ll notice the inequalities’ graphs never intersect, confirming there's no solution. Visualizing inequalities this way makes inconsistencies and feasibilities of solutions clear at a glance.