Problem 93 - Algebra for College Students

Question

Solve each equation. $$ \left|x-\frac{1}{2}\right|=\left|\frac{1}{2} x-2\right| $$

Step-by-Step Solution

Answer

The solutions are \(x = -3\) and \(x = \frac{5}{3}\).

Step 1 - Understand the absolute value equation

The equation \ \ \(\left|x-\frac{1}{2}\right|=\left|\frac{1}{2}x-2\right|\) involves absolute values on both sides. We need to solve for the values of \(x\) which make the expressions inside the absolute values equal in magnitude, but potentially different in sign.

Step 2 - Set up two cases for the equality

Since absolute values represent the distance from zero and can be positive or negative, split the original equation into two separate cases:Case 1: \ \ \(x-\frac{1}{2}=\frac{1}{2}x-2\)Case 2: \ \ \(x-\frac{1}{2}=-(\frac{1}{2}x-2)\)

Step 3 - Solve Case 1 equation

Solve the first case by isolating \(x\):\ \ \(x-\frac{1}{2}=\frac{1}{2}x-2\)Subtract \(\frac{1}{2}x\) from both sides:\ \ \(x-\frac{1}{2}-\frac{1}{2}x= -2\)Combine like terms:\ \ \(\frac{1}{2}x-\frac{1}{2}= -2\)Add \(\frac{1}{2}\) to both sides:\ \ \(\frac{1}{2}x=-\frac{3}{2}\)Multiply by 2 to solve for \(x\):\ \ \(x=-3\)

Step 4 - Solve Case 2 equation

Solve the second case by isolating \(x\):\ \ \(x-\frac{1}{2}=-(\frac{1}{2}x-2)\)Distribute the negative sign on the right-hand side:\ \ \(x-\frac{1}{2}=-\frac{1}{2}x+2\)Add \(\frac{1}{2}x\) to both sides:\ \ \(x+\frac{1}{2}x-\frac{1}{2}=2\)Combine like terms:\ \ \(\frac{3}{2}x-\frac{1}{2}=2\)Add \(\frac{1}{2}\) to both sides:\ \ \(\frac{3}{2}x=2.5\)Multiply by \(\frac{2}{3}\) to solve for \(x\):\ \ \(x=\frac{5}{3}\)

Step 5 - Verify solutions

Substitute \(x=-3\) and \(x=\frac{5}{3}\) back into the original equation to verify. Both should satisfy the original equation, hence they are solutions.

Key Concepts

absolute value properties

The absolute value of a number, denoted as \(|a|\), represents the distance of that number from zero on the number line, regardless of direction. This means that both \(|5|\) and \(|-5|\) equal 5 because distance is always positive. In equations, absolute values create scenarios where you need to consider both the positive and negative cases of the numbers inside the absolute value bars.
Let's understand this with an example: \(|x - 2| = 3\). This can be rephrased as two separate equations: \(x - 2 = 3\) and \(x - 2 = -3\). Understanding this property is crucial when dealing with absolute value equations, like the one in our exercise.

equation solving techniques

Solving absolute value equations involves a few key steps. First, rewrite the equation to account for the fact that the absolute value represents both the positive and negative cases.
For the given equation, \(|x - \frac{1}{2}| = |\frac{1}{2}x - 2|\), we split it into two cases based on equality:

• Case 1: \(x - \frac{1}{2} = \frac{1}{2}x - 2\)
• Case 2: \(x - \frac{1}{2} = - (\frac{1}{2}x - 2)\)

Once split, solve each equation separately. Combine like terms, isolate the variable, and solve step by step.
In our example, for Case 1:

• Simplify: \(x - \frac{1}{2} = \frac{1}{2}x - 2\)
• Subtract \(\frac{1}{2}x\) from both sides: \(x - \frac{1}{2} - \frac{1}{2}x = -2\)


• Combine like terms: \(\frac{1}{2}x - \frac{1}{2} = -2\)


• Add \(\frac{1}{2}\) to both sides: \(\frac{1}{2}x = -\frac{3}{2}\)


• Multiply by 2 to solve for \(|3\)
• Similarly for Case 2, distribution and stepwise addition/subtraction will get: \(x = \frac{5}{3}\).

This step-by-step process ensures you don't miss any detail when solving absolute value equations.

case analysis in algebra

Case analysis is a powerful technique in algebra, particularly with absolute value equations. When faced with \(|x - \frac{1}{2}| = |\frac{1}{2}x - 2|\), you realize that both sides of the equation can equal either a positive or negative result.
Therefore, you create two instances:

• Case 1 investigates when \(x - \frac{1}{2}\) equals \(\frac{1}{2}x - 2\).


• Case 2 explores when \(x - \frac{1}{2}\) equals \(- (\frac{1}{2}x - 2)\).

Splitting the problem in this manner enables a thorough analysis.
For each case, proceed as follows:

• Combine like terms
• Isolate the variable
• Solve for the unknown.

This method ensures that you consider all possible solutions and thoroughly check the behavior of the equation under different scenarios.
Finally, always substitute the solutions back into the original equation to verify their correctness. This prevents errors and ensures the solutions are valid in the original equation's context, as seen when we validated \(x=-3\) and \(x=\frac{5}{3}\) in our problem.