Problem 6 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 10, solve for \(x\). $$ |x-2|=|3-2 x| $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{5}{3} \) and \( x = 1 \).

Step 1: Identify Absolute Values

The given equation is \[|x-2|=|3-2x| \]. To solve this, consider the properties of absolute values. There are two cases to consider: when the expressions inside the absolute value are non-negative and when they are negative.

Step 2: Case 1 - Both Expressions Non-Negative

Assume both expressions inside the absolute values are non-negative, meaning \[ x-2 = 3-2x \]. Solve for \( x \) by first adding \( 2x \) to both sides of the equation: \[ x + 2x - 2 = 3 \] Simplify this to \[ 3x - 2 = 3 \]. Then, add 2 to both sides: \[ 3x = 5 \] Divide both sides by 3: \[ x = \frac{5}{3} \].

Step 3: Case 2 - Both Expressions Negative

Assume both expressions are negative, meaning \[ x-2 = -(3-2x) \]. Simplify the right side: \[ x - 2 = -3 + 2x \]. Move all terms involving \( x \) to one side by subtracting \( x \) from both sides: \[ -2 = -3 + x \]. Add 3 to both sides: \[ 1 = x \].

Step 4: Verification

Verify the solutions \( x = \frac{5}{3} \) and \( x = 1 \) in the original equation. For \( x = \frac{5}{3} \): \[ | \frac{5}{3} - 2 | = | 3 - 2 \times \frac{5}{3} | \] Calculation confirms both sides equal \( \frac{1}{3} \). For \( x = 1 \): \[ |1-2| = |3-2 \cdot 1| \] Calculation confirms both sides equal 1.

Key Concepts

Solving equations

To solve equations involving absolute values, we need to break them down into simpler cases. The absolute value symbol \(| \cdot | \) essentially means that the value inside can be either positive or negative. This creates multiple cases that we should consider.
For example, consider the equation given in the exercise: \(|x-2|=|3-2x| \). This represents an equation where the expressions inside absolute values could have different signs.
Therefore, we will solve the original equation by separating it into different cases based on the properties of the absolute value.

• Case 1: Both expressions inside the absolute values are non-negative.
• Case 2: Both expressions inside the absolute values are non-positive.

It's essential to simplify each case to find the value of the variable x. Let’s look at each case step by step closely.

Absolute value properties

The absolute value of a number represents its distance from zero on a number line, regardless of direction. In mathematical terms, for any real number \(\text{a}\), the absolute value is denoted by \(|\text{a}|\), and it has the following properties:

• \(| \text{a} | \geq 0\)
• \(|-\text{a}| = |\text{a}| \)
• \(| \text{a} \times \text{b} | = |\text{a}| \times |\text{b}| \)
• \(| \text{a} / \text{b} | = |\text{a}| / |\text{b}| \), where \(\text{b} \e 0\)

When working with equations involving absolute values, these properties help us understand how to split the equation into simpler cases, each where the expression inside the absolute value is set equal to both positive and negative versions of the other side.
Let’s use these principles, focusing on the exercise:

• The first case is where both expressions are non-negative, so we can solve \(\text{x-2} = \text{3-2x}\).
• The second case is where both expressions might involve a negative, thus solving \(\text{x-2} = -(\text{3-2x})\).

For each case, isolate the variable and simplify the resulting equation.

Verification of solutions

After solving the equations from each case, it’s important to verify the solutions. This means plugging the solutions back into the original equation to make sure they satisfy it.
For the exercise given:

• We found the solutions \(\text{x = } \frac{5}{3}\) and \(\text{x = 1}\).

We need to check if these solutions make both sides of the original equation equal:

• For \(\text{x = } \frac{5}{3}\):
  
  • Calculate \(|\frac{5}{3} - 2|\) and \(|3 - 2 \times \frac{5}{3}|\).
  • Both sides should equal \(\frac{1}{3}\).
• For \(\text{x = 1}\):
  
  • Calculate \(|1-2|\) and \(|3-2 \times 1|\).
  • Both sides should equal 1.

If both solutions satisfy the original equation, we can confirm them as correct. This step is crucial to ensure no mistakes were made during the solutions process and each possible solution is valid.