Problem 26 - College Algebra with Modeling and Visualization

Question

Solve the inequality symbolically. Express the solution set in set-builder or interval notation. $$ -5<1-2 x<40 $$

Step-by-Step Solution

Answer

The solution is \((-19.5, 3)\).

Step 1: Interpret the Compound Inequality

The compound inequality \(-5 < 1 - 2x < 40\) is actually two inequalities combined: \(-5 < 1 - 2x\) and \(1 - 2x < 40\). We need to solve each inequality separately and then find the intersection of the solutions to get the final answer.

Step 2: Solve the First Inequality

Solve for \(x\) in the inequality \(-5 < 1 - 2x\).First, subtract 1 from both sides:\[-5 - 1 < -2x\]\[-6 < -2x\]Next, divide both sides by \(-2\) and remember to flip the inequality sign:\[3 > x\] or equivalently \(x < 3\).

Step 3: Solve the Second Inequality

Now, solve for \(x\) in the inequality \(1 - 2x < 40\).First, subtract 1 from both sides:\[1 - 1 - 2x < 40 - 1\]\[-2x < 39\]Next, divide both sides by \(-2\) and remember to flip the inequality sign:\[x > -\frac{39}{2}\] or equivalently \(x > -19.5\).

Step 4: Find the Intersection of Solutions

The solution to the compound inequality is the intersection of the solutions from Step 2 and Step 3. Thus, we have:\(-19.5 < x < 3\).

Step 5: Express the Solution in Interval Notation

The solution \(-19.5 < x < 3\) can be expressed in interval notation as \((-19.5, 3)\). This interval represents all the values of \(x\) that satisfy both parts of the compound inequality.

Key Concepts

Compound Inequalities

Compound inequalities involve solving two or more inequalities that are combined into one statement by the word "and" or "or". These inequalities are crucial because they allow us to find solutions that satisfy multiple conditions at once. For the compound inequality

• \(-5 < 1 - 2x < 40\),

we actually have two separate inequalities:

• \(-5 < 1 - 2x\)
• \(1 - 2x < 40\).

These need to be solved individually to determine where their solution sets overlap. Solving each inequality often involves basic algebraic methods such as addition, subtraction, multiplication, or division. Remember, if you multiply or divide by a negative number, you must flip the inequality sign. Once solved, the solutions of individual inequalities are combined. For these specific inequalities, the combined solution is the intersection of the two sets, which indicates the values that satisfy both inequalities at the same time. It's like having two rules, and finding what numbers obey both of them simultaneously.

Set-Builder Notation

Set-builder notation is a concise way of expressing a set by specifying a property that its members must satisfy. This notation is particularly valuable when describing complex solution sets in algebra. It describes the elements of a set rather than listing them. Consider the solution we found for the compound inequality:
\(-19.5 < x < 3\).
In set-builder notation, this solution can be expressed as:

• \(\{ x \mid -19.5 < x < 3 \}\).

The vertical bar, "\(\mid\)", is read as "such that", indicating that the set includes all values of \(x\) that satisfy the condition presented. With set-builder notation, we highlight the defining condition or property of the elements within the set, helping clarify the scope of solutions in a mathematical expression.

Interval Notation

Interval notation is a streamlined way of describing the set of solutions to inequalities. It uses parentheses and brackets to indicate where intervals start and stop, showing which endpoints are included or excluded. For our compound inequality solution,
\(-19.5 < x < 3\),
we express this in interval notation as

• \((-19.5, 3)\).

In this representation, parentheses \(()\) indicate that the endpoints are not included in the set (i.e., \(x\) can be any number greater than \(-19.5\) and less than \(3\)). If an endpoint were included, we would use a bracket \([]\). Interval notation is efficient for its brevity and visual clarity. By using interval notation, we convey compactly where the solutions to an inequality begin and end, without explicitly listing out all the possible solutions, perfect for continuous ranges of values.