The equation |2x - 3| = x - 5 has how many solutions for x?

1. Translate the problem requirements: We need to find how many values of x satisfy the equation \(|2\mathrm{x} - 3| = \mathrm{x} - 5\). Solutions means values of x that make both sides equal.
2. Identify the critical point for the absolute value: Determine when the expression inside the absolute value changes sign to set up our cases.
3. Set up and solve two separate cases: Handle the positive and negative cases of the absolute value expression separately.
4. Validate solutions against original constraints: Check if our solutions actually work in the original equation and satisfy the conditions for each case.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have the equation \(|2\mathrm{x} - 3| = \mathrm{x} - 5\), and we need to count how many different values of x make this equation true.

Think of it this way: we're looking for x-values where the absolute value expression on the left equals the linear expression on the right. The absolute value bars mean "distance from zero," so \(|2\mathrm{x} - 3|\) represents the distance of \((2\mathrm{x} - 3)\) from zero.

Process Skill: TRANSLATE

2. Identify the critical point for the absolute value

To handle absolute value equations, we need to figure out when the expression inside the bars changes from positive to negative. This happens when \(2\mathrm{x} - 3 = 0\).

Solving: \(2\mathrm{x} - 3 = 0\)
\(2\mathrm{x} = 3\)
\(\mathrm{x} = \frac{3}{2} = 1.5\)

So at \(\mathrm{x} = 1.5\), the expression \((2\mathrm{x} - 3)\) equals zero. When \(\mathrm{x} > 1.5\), the expression \((2\mathrm{x} - 3)\) is positive. When \(\mathrm{x} < 1.5\), the expression \((2\mathrm{x} - 3)\) is negative.

This critical point divides our problem into two separate cases to consider.

3. Set up and solve two separate cases

Case 1: When \(\mathrm{x} \geq 1.5\)
When \(\mathrm{x} \geq 1.5\), we have \(2\mathrm{x} - 3 \geq 0\), so \(|2\mathrm{x} - 3| = 2\mathrm{x} - 3\)

Our equation becomes: \(2\mathrm{x} - 3 = \mathrm{x} - 5\)
Solving: \(2\mathrm{x} - \mathrm{x} = -5 + 3\)
\(\mathrm{x} = -2\)

But wait! We assumed \(\mathrm{x} \geq 1.5\) for this case, and we got \(\mathrm{x} = -2\). Since \(-2 < 1.5\), this solution doesn't satisfy our case condition. So Case 1 gives us no valid solution.

Case 2: When \(\mathrm{x} < 1.5\)
When \(\mathrm{x} < 1.5\), we have \(2\mathrm{x} - 3 < 0\), so \(|2\mathrm{x} - 3| = -(2\mathrm{x} - 3) = -2\mathrm{x} + 3\)

Our equation becomes: \(-2\mathrm{x} + 3 = \mathrm{x} - 5\)
Solving: \(-2\mathrm{x} - \mathrm{x} = -5 - 3\)
\(-3\mathrm{x} = -8\)
\(\mathrm{x} = \frac{8}{3} \approx 2.67\)

But we assumed \(\mathrm{x} < 1.5\) for this case, and we got \(\mathrm{x} = \frac{8}{3} \approx 2.67\). Since \(2.67 > 1.5\), this solution doesn't satisfy our case condition either. So Case 2 also gives us no valid solution.

Process Skill: CONSIDER ALL CASES

4. Validate solutions against original constraints

Let's double-check by considering what we found:

• Case 1 \((\mathrm{x} \geq 1.5)\): We got \(\mathrm{x} = -2\), but \(-2 < 1.5\), so this is invalid
• Case 2 \((\mathrm{x} < 1.5)\): We got \(\mathrm{x} = \frac{8}{3}\), but \(\frac{8}{3} > 1.5\), so this is invalid

We should also check if there are any other constraints. For the original equation \(|2\mathrm{x} - 3| = \mathrm{x} - 5\) to make sense, we need the right side to be non-negative (since absolute values are always non-negative).

So we need: \(\mathrm{x} - 5 \geq 0\), which means \(\mathrm{x} \geq 5\)

This additional constraint makes it even clearer why we have no solutions. Even if our case solutions were valid, neither \(-2\) nor \(\frac{8}{3}\) satisfies \(\mathrm{x} \geq 5\).

Process Skill: APPLY CONSTRAINTS

5. Final Answer

Since neither case produced a valid solution that satisfies all the necessary conditions, the equation \(|2\mathrm{x} - 3| = \mathrm{x} - 5\) has 0 solutions.

The answer is A: 0.

Common Faltering Points

Errors while devising the approach

1. Missing the non-negativity constraint: Students often forget that since \(|2\mathrm{x} - 3| \geq 0\) always, we must have \(\mathrm{x} - 5 \geq 0\), which means \(\mathrm{x} \geq 5\). This is a crucial constraint that eliminates many potential solutions, but students frequently overlook it when setting up their approach.

2. Incorrectly handling the absolute value setup: Students may struggle to correctly identify that when \(2\mathrm{x} - 3 < 0\), the absolute value \(|2\mathrm{x} - 3|\) equals \(-(2\mathrm{x} - 3) = -2\mathrm{x} + 3\). They might incorrectly write it as \(2\mathrm{x} - 3\) or make sign errors in the transformation.

Errors while executing the approach

1. Failing to check case validity: After solving each case algebraically, students often forget to verify whether their solution actually satisfies the original case condition. For example, getting \(\mathrm{x} = -2\) in Case 1 but not checking that \(-2 < 1.5\) violates the assumption \(\mathrm{x} \geq 1.5\).

2. Arithmetic errors in case solving: When solving \(-2\mathrm{x} + 3 = \mathrm{x} - 5\) in Case 2, students may make sign errors or incorrectly combine like terms, potentially getting \(-3\mathrm{x} = -8\) wrong or miscalculating \(\mathrm{x} = \frac{8}{3}\).

Errors while selecting the answer

1. Accepting invalid solutions: Even after finding that neither case produces a valid solution, students might still choose answer choice B (1 solution) or C (2 solutions) by incorrectly counting the algebraic solutions \(\mathrm{x} = -2\) and \(\mathrm{x} = \frac{8}{3}\) without recognizing they don't satisfy the necessary constraints.