2x + y = 12|y| leq 12 For how many ordered pairs (x, y) that are solutions of the system...

1. Translate the problem requirements: We need to find ordered pairs (x, y) where both x and y are integers that satisfy both conditions: \(2\mathrm{x} + \mathrm{y} = 12\) and \(|\mathrm{y}| ≤ 12\) (meaning y is between -12 and 12, inclusive)
2. Express x in terms of y using the linear equation: Rearrange \(2\mathrm{x} + \mathrm{y} = 12\) to get \(\mathrm{x} = \frac{12 - \mathrm{y}}{2}\), which shows that for x to be an integer, \((12 - \mathrm{y})\) must be even
3. Determine which y-values yield integer x-values: Since 12 is even, \((12 - \mathrm{y})\) is even only when y is even, so we need to find all even integers y where \(|\mathrm{y}| ≤ 12\)
4. Count all valid even y-values within the constraint: List all even integers from -12 to 12 inclusive and count them to get the total number of ordered pairs

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in plain English. We have two conditions that must both be satisfied:

First condition: The equation \(2\mathrm{x} + \mathrm{y} = 12\) tells us that if we take twice the x-value and add the y-value, we get 12.

Second condition: \(|\mathrm{y}| ≤ 12\) means that y can be any value from -12 to +12, including both endpoints.

We need both x and y to be integers (whole numbers like ..., -2, -1, 0, 1, 2, ...), and we want to count how many such ordered pairs (x, y) exist.

Process Skill: TRANSLATE - Converting the mathematical notation into clear requirements

2. Express x in terms of y using the linear equation

Now let's figure out the relationship between x and y. From the equation \(2\mathrm{x} + \mathrm{y} = 12\), we can solve for x:

Starting with: \(2\mathrm{x} + \mathrm{y} = 12\)
Subtract y from both sides: \(2\mathrm{x} = 12 - \mathrm{y}\)
Divide both sides by 2: \(\mathrm{x} = \frac{12 - \mathrm{y}}{2}\)

This tells us that once we pick a value for y, the value of x is completely determined. But here's the key insight: for x to be an integer, the expression \((12 - \mathrm{y})\) must be divisible by 2, which means \((12 - \mathrm{y})\) must be even.

3. Determine which y-values yield integer x-values

Let's think about when \((12 - \mathrm{y})\) is even. Since 12 is an even number, the difference \((12 - \mathrm{y})\) will be even only when y is also even.

Here's why: even - even = even, but even - odd = odd

So we need y to be an even integer, and we also need \(|\mathrm{y}| ≤ 12\).

Let's test this with a few examples:
- If y = 0 (even): \(\mathrm{x} = \frac{12 - 0}{2} = 6\) ✓ (integer)
- If y = 2 (even): \(\mathrm{x} = \frac{12 - 2}{2} = 5\) ✓ (integer)
- If y = 1 (odd): \(\mathrm{x} = \frac{12 - 1}{2} = 5.5\) ✗ (not an integer)
- If y = -4 (even): \(\mathrm{x} = \frac{12 - (-4)}{2} = 8\) ✓ (integer)

Process Skill: INFER - Recognizing the even/odd pattern that determines valid solutions

4. Count all valid even y-values within the constraint

Now we need to find all even integers where \(|\mathrm{y}| ≤ 12\), which means \(-12 ≤ \mathrm{y} ≤ 12\).

The even integers in this range are:
-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12

Let's count them systematically:
- Negative even integers: -12, -10, -8, -6, -4, -2 (that's 6 values)
- Zero: 0 (that's 1 value)
- Positive even integers: 2, 4, 6, 8, 10, 12 (that's 6 values)

Total count: 6 + 1 + 6 = 13

Let's verify with a couple more examples:
- y = 12: \(\mathrm{x} = \frac{12 - 12}{2} = 0\), so (0, 12) works
- y = -12: \(\mathrm{x} = \frac{12 - (-12)}{2} = 12\), so (12, -12) works

4. Final Answer

We found exactly 13 ordered pairs (x, y) where both x and y are integers and satisfy both conditions.

The answer is (D) 13.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the absolute value constraint: Students often misinterpret \(|\mathrm{y}| ≤ 12\) as only considering positive values of y \((0 ≤ \mathrm{y} ≤ 12)\), forgetting that absolute value includes both positive and negative values. This leads them to only consider half the valid y-values.

2. Not recognizing the integer constraint for x: Students may see that y must be an integer but fail to realize that for x to also be an integer, the expression \(\frac{12 - \mathrm{y}}{2}\) must yield a whole number. This means they don't identify the even/odd pattern requirement for y.

3. Treating the problem as finding y first vs. x first: Some students attempt to find integer values of x first and then solve for y, which makes the problem more complex. The more efficient approach is to use the constraint on y to determine valid values.

Errors while executing the approach

1. Arithmetic errors when calculating x values: When computing \(\mathrm{x} = \frac{12 - \mathrm{y}}{2}\) for various y values, students may make simple calculation mistakes, especially with negative y values. For example, calculating \(\frac{12 - (-10)}{2}\) incorrectly as 1 instead of 11.

2. Missing boundary values: Students often forget to include the boundary cases y = 12 and y = -12 in their count, thinking these extreme values might not be valid solutions.

3. Incorrect identification of even integers: Some students may incorrectly list the even integers in the range \(-12 ≤ \mathrm{y} ≤ 12\), either by including odd numbers or missing some even numbers, particularly around zero or the negative values.

Errors while selecting the answer

1. Miscounting the total valid pairs: Even after correctly identifying all valid y values, students may miscount when adding up the negative even integers (6) + zero (1) + positive even integers (6), arriving at 12 instead of 13.

2. Double-counting or under-counting boundary cases: Students might count y = 0 twice (once as non-negative and once as non-positive) or forget to count it entirely, leading to incorrect final totals.