Problem 31 - Precalculus Mathematics for Calculus

Question

Solve the linear inequality. Express the solution using interval notation and graph the solution set. $$-2<8-2 x \leq-1$$

Step-by-Step Solution

Answer

The solution is \([\frac{9}{2}, 5)\].

Step 1: Break Down the Compound Inequality

The given inequality is compound: - \(-2 < 8 - 2x\) and \(8 - 2x \leq -1\). We need to solve these two inequalities separately to find the solution.

Step 2: Solve the First Inequality

For the first part: \(-2 < 8 - 2x\)Subtract 8 from both sides: \(-2 - 8 < -2x\)This simplifies to:\(-10 < -2x\)Divide both sides by -2, remembering to reverse the inequality sign:\(5 > x\) or equivalently, \(x < 5\).

Step 3: Solve the Second Inequality

For the second part: \(8 - 2x \leq -1\)Subtract 8 from both sides: \(-2x \leq -1 - 8\)This simplifies to:\(-2x \leq -9\)Divide both sides by -2, remembering to reverse the inequality sign:\(x \geq \frac{9}{2}\).

Step 4: Combine the Solutions

The solutions from Steps 2 and 3 are \(x < 5\) and \(x \geq \frac{9}{2}\). Combined, the solution in interval notation is:\[\left[\frac{9}{2}, 5\right)\].

Step 5: Graph the Solution

On a number line, mark the points \(\frac{9}{2}\) and \(5\).Use a closed circle at \(\frac{9}{2}\) to indicate \(x \geq \frac{9}{2}\) and an open circle at \(5\) to indicate \(x < 5\). Shade the region between \(\frac{9}{2}\) and \(5\). This is the solution set.

Key Concepts

Interval Notation

Interval notation is a concise way to express the set of solutions for inequalities, showing exactly where a variable lies on the number line. In this case, for the compound inequality

• \(-2 < 8 - 2x \leq -1\)

we found that the solutions are for values of \(x\) such that \(x \geq \frac{9}{2}\) and \(x < 5\). When expressed in interval notation, these solutions convert to:

• \[\left[\frac{9}{2}, 5\right)\]

The square bracket \([\frac{9}{2}\)) denotes that \(\frac{9}{2}\) is included in the solution set, meaning \(x\) can be equal to this value. The parenthesis \((5)\) indicates that 5 is not included, meaning \(x\) can approach but not equal 5.
This notation offers a clear and compact representation of the solution region on the number line.
By becoming familiar with interval notation, you can efficiently communicate intervals and open/closed points of inequalities without excessively long descriptions.

Compound Inequalities

Compound inequalities involve two distinct inequalities that are combined into one statement by the words "and" or "or."
For this exercise, the compound inequality is

• \(-2 < 8 - 2x\)
• \(8 - 2x \leq -1\)

The use of "less than" and "less than or equal to" allows for combining the solutions from each inequality.
When solving compound inequalities:

• Solve each inequality separately.
• Identify the solution sets for each part.
• Combine the solution sets based on whether the statement is "and" (intersection of solutions) or "or" (union of solutions).

In this context, "and" indicates the intersection, meaning both conditions must be true simultaneously.
Therefore, the shared solutions make up the set where both conditions overlap, leading to an intersection represented by the interval \[\left[\frac{9}{2}, 5\right)\].

Solution Set

The solution set of a linear inequality or a compound inequality expresses all possible values that satisfy the condition or conditions presented by the inequalities. For

• \(-2 < 8 - 2x \leq -1\)

we solved the inequality and found that all values of \(x\) which satisfy both parts are captured in the interval:

• \[\left[\frac{9}{2}, 5\right)\]

The solution set is visualized on the number line, which helps in understanding the set of values \(x\) can take:

• \(\frac{9}{2}\) is part of the solution (indicated by a filled circle);
• 5 is not part of the solution (indicated by an open circle).

Graphically marking the solution set involves shading the region between these two points, representing all the permissible values of \(x\).
Identifying and understanding how to draw solution sets is crucial in determining the range of permissible values that satisfy the given inequalities.