Problem 90 - Algebra for College Students

Question

Solve each equation or inequality. $$ \left|\frac{2}{3} x+\frac{1}{6}\right|+\frac{1}{2}=\frac{5}{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{11}{4}\) and \(-\frac{13}{4}\).

Step 1 - Isolate the Absolute Value Expression

First, subtract \(\frac{1}{2}\) from both sides of the equation to isolate the absolute value expression.\[ \left|\frac{2}{3} x+\frac{1}{6}\right| = \frac{5}{2} - \frac{1}{2} \] Simplifying the right side, we get: \[\frac{2}{3} x + \frac{1}{6} = 2 \]

Step 2 - Eliminate the Absolute Value

Solve the equation \(\left|\frac{2}{3} x+\frac{1}{6}\right| = 2\) by considering the positive and negative cases. The positive case is: \[\frac{2}{3} x + \frac{1}{6} = 2 \] The negative case is: \[\frac{2}{3} x + \frac{1}{6} = -2 \]

Step 3 - Solve the Positive Case

Solve for \(x\) in the positive case: \[\frac{2}{3} x + \frac{1}{6} = 2 \] Subtract \(\frac{1}{6}\) from both sides: \[\frac{2}{3} x = 2 - \frac{1}{6} \] Simplify: \[\frac{2}{3} x = \frac{12}{6} - \frac{1}{6} = \frac{11}{6} \] Multiply both sides by \(\frac{3}{2}\): \[\frac{2}{3} x \cdot \frac{3}{2} = \frac{11}{6} \cdot \frac{3}{2} \] Simplifying: \[\boxed{x = \frac{11}{4}} \]

Step 4 - Solve the Negative Case

Solve for \(x\) in the negative case: \[\frac{2}{3} x + \frac{1}{6} = -2 \] Subtract \(\frac{1}{6}\) from both sides: \[\frac{2}{3} x = -2 - \frac{1}{6} \] Simplify: \[\frac{2}{3} x = -\frac{12}{6} - \frac{1}{6} = -\frac{13}{6} \] Multiply both sides by \(\frac{3}{2}\): \[\frac{2}{3} x \cdot \frac{3}{2} = -\frac{13}{6} \cdot \frac{3}{2} \] Simplifying: \[\boxed{x = -\frac{13}{4}} \]

Key Concepts

Isolating the Absolute Value

The first step in solving an absolute value equation is to isolate the absolute value expression on one side of the equation. We need to move all other terms to the opposite side. Let's look at the given equation: \[ \frac{2}{3} x + \frac{1}{6} + \frac{1}{2} = \frac{5}{2} \].
To isolate the absolute value, subtract \( \frac{1}{2} \) from both sides of the equation:
\[ \frac{2}{3} x + \frac{1}{6} = \frac{5}{2} - \frac{1}{2} \]. Simplifying the right side gives \( \frac{4}{2} = 2 \). Now, the equation looks like:
\[ \frac{2}{3} x + \frac{1}{6} = 2 \].

Positive and Negative Cases

For any absolute value equation \( \left| A \right| = B \), we set up two cases to remove the absolute value: the positive case \( A = B \) and the negative case \( A = -B \). Here, \( A = \frac{2}{3} x + \frac{1}{6} \) and \( B = 2 \) from our isolated absolute value expression.

For the positive case, we get: \[ \frac{2}{3} x + \frac{1}{6} = 2 \].

For the negative case, we get:\[ \frac{2}{3} x + \frac{1}{6} = -2 \].
These cases allow us to treat the equation as a regular linear equation for each scenario.

Solving Linear Equations

To solve for \( x \) in both cases, we need to follow the steps for solving linear equations.

In the positive case:
\[ \frac{2}{3} x + \frac{1}{6} = 2 \]. Subtract \( \frac{1}{6} \) from both sides: \[ \frac{2}{3} x = \frac{11}{6} \].
Multiply both sides by \( \frac{3}{2} \) to isolate \( x \):
\[ x = \frac{11}{4} \].
In the negative case:

\[ \frac{2}{3} x + \frac{1}{6} = -2 \]. Subtract \( \frac{1}{6} \) from both sides: \[ \frac{2}{3} x = -\frac{13}{6} \].
Multiply both sides by \( \frac{3}{2} \) to isolate \( x \): \[ x = -\frac{13}{4} \].
Applying these steps, we've successfully solved for \( x \) in both scenarios.