Problem 18 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 26 , find the solution set of the given inequality, and illustrate the solution on the real number line. $$ |6-2 x| \geq 7 $$

Step-by-Step Solution

Answer

Solution set: \(x \leq -\frac{1}{2}\) or \(x \geq 6.5\).

Step 1: Understand the Absolute Value Inequality

The given inequality is \(| 6 - 2x | \geq 7\). This means that the distance between \(6 - 2x\) and zero is greater than or equal to \7\.

Step 2: Break Down the Absolute Value Inequality

To solve \(| A | \geq B\), we rewrite it as two separate inequalities: \( A \geq B\) and \( A \leq -B\). For our case, let \(A = 6 - 2x\) and \(B = 7\). This gives us: \((6 - 2x \geq 7)\) and \((6 - 2x \leq -7)\).

Step 3: Solve the First Inequality

Solve \(6 - 2x \geq 7\). Subtract \6\ from both sides: \(-2x \geq 1\). Divide both sides by \(-2\) and reverse the inequality sign: \(x \leq -\frac{1}{2}\).

Step 4: Solve the Second Inequality

Solve \(6 - 2x \leq -7\). Subtract \6\ from both sides: \(-2x \leq -13\). Divide both sides by \(-2\) and reverse the inequality sign: \(x \geq \frac{13}{2}\) or \(x \geq 6.5\).

Step 5: Combine the Solution Sets

Combine the solutions from Steps 3 and 4. The solution set is \(x \leq -\frac{1}{2}\) or \(x \geq \frac{13}{2}\).

Step 6: Illustrate on the Real Number Line

Draw a number line and shade the regions corresponding to the solution set. Shade from negative infinity to \(-\frac{1}{2}\) and from \(\frac{13}{2}\) to positive infinity, including \(-\frac{1}{2}\) and \(\frac{13}{2}\).

Key Concepts

Solving Inequalities

Solving inequalities involves finding the range of values that satisfy a given condition. In this case, we are dealing with an absolute value inequality. To solve it, we break it down into two simpler inequalities. When you have \(|A| \geq B\), it translates to \(A \geq B\) and \(A \leq -B\). This means we need to find where these conditions hold true.

By solving each inequality separately, we obtain two distinct solutions. These solutions will then form the full answer. Combining these solutions and representing them on the real number line completes the process. Remember, always to reverse the inequality sign when dividing by a negative number.

Real Number Line

The real number line is a visual representation of all possible real numbers. When illustrating solutions on this line, each solution or range of solutions is marked appropriately. For the given inequality \(|6-2x| \geq 7\), we found that solutions are \(x \leq -\frac{1}{2}\) and \(x \geq 6.5\).

On the real number line, this means shading two separate regions:

• From negative infinity to \(-\frac{1}{2}\), including \(-\frac{1}{2}\)
• From \(6.5\) to positive infinity, including \(6.5\)

This visual helps you see where the solutions to the inequality lie. It separates the number line into regions that do and do not satisfy the inequality.

Breakdown of Absolute Value

Absolute value represents the distance of a number from zero on the number line, regardless of direction. In this case, the inequality \(|6-2x| \geq 7\) tells us that the expression \(6-2x\) must be at least 7 units away from zero.

To unpack this, we rewrite it as two inequalities:

\(6-2x \geq 7\) \(6-2x \leq -7\)

Each inequality is then solved step-by-step, ensuring we handle the algebra correctly. For the first, rearranging yields \(x \leq -\frac{1}{2}\). For the second, we get \(x \geq 6.5\). These inequalities show where \(6-2x\) is sufficiently far from zero, illustrating the concept of absolute value in inequalities.