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 all ordered pairs \(\mathrm{x}, \mathrm{y}\) where both x and y are integers, y satisfies \(|\mathrm{y}| \leq 12\), and the pair satisfies \(2\mathrm{x} + \mathrm{y} = 12\)
2. Identify the constraint range for y: Determine all possible integer values of y from the absolute value constraint
3. Calculate corresponding x values: For each valid y value, use the linear equation to find the corresponding x value
4. Verify integer solutions: Check that each calculated x value is indeed an integer, and count all valid pairs

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday language. We have two conditions that must be satisfied simultaneously:

First, we have the equation \(2\mathrm{x} + \mathrm{y} = 12\). This means that when we multiply x by 2 and add y, we get 12.

Second, we have the constraint \(|\mathrm{y}| \leq 12\), which means the absolute value of y is at most 12. In plain English, this means y can be anywhere from -12 to +12, including both endpoints.

We need both x and y to be integers (whole numbers, including negative numbers and zero), and we want to count how many different ordered pairs \(\mathrm{x}, \mathrm{y}\) satisfy both conditions.

Process Skill: TRANSLATE - Converting the mathematical constraints into clear requirements we can work with

2. Identify the constraint range for y

Since \(|\mathrm{y}| \leq 12\), let's think about what values y can take.

The absolute value constraint means:

• y can be as small as -12
• y can be as large as +12
• y can be any integer in between

So the possible integer values for y are: -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

That gives us 25 possible values for y (from -12 to +12 inclusive).

3. Calculate corresponding x values

Now, for each possible value of y, we need to find the corresponding x value using the equation \(2\mathrm{x} + \mathrm{y} = 12\).

We can rearrange this equation to solve for x: \(2\mathrm{x} = 12 - \mathrm{y}\), so \(\mathrm{x} = \frac{12 - \mathrm{y}}{2}\)

Let's work through each y value:

• When \(\mathrm{y} = -12\): \(\mathrm{x} = \frac{12 - (-12)}{2} = \frac{24}{2} = 12\) ✓ (integer)
• When \(\mathrm{y} = -11\): \(\mathrm{x} = \frac{12 - (-11)}{2} = \frac{23}{2} = 11.5\) ✗ (not integer)
• When \(\mathrm{y} = -10\): \(\mathrm{x} = \frac{12 - (-10)}{2} = \frac{22}{2} = 11\) ✓ (integer)
• When \(\mathrm{y} = -9\): \(\mathrm{x} = \frac{12 - (-9)}{2} = \frac{21}{2} = 10.5\) ✗ (not integer)
• When \(\mathrm{y} = -8\): \(\mathrm{x} = \frac{12 - (-8)}{2} = \frac{20}{2} = 10\) ✓ (integer)
• When \(\mathrm{y} = -7\): \(\mathrm{x} = \frac{12 - (-7)}{2} = \frac{19}{2} = 9.5\) ✗ (not integer)
• When \(\mathrm{y} = -6\): \(\mathrm{x} = \frac{12 - (-6)}{2} = \frac{18}{2} = 9\) ✓ (integer)
• When \(\mathrm{y} = -5\): \(\mathrm{x} = \frac{12 - (-5)}{2} = \frac{17}{2} = 8.5\) ✗ (not integer)
• When \(\mathrm{y} = -4\): \(\mathrm{x} = \frac{12 - (-4)}{2} = \frac{16}{2} = 8\) ✓ (integer)
• When \(\mathrm{y} = -3\): \(\mathrm{x} = \frac{12 - (-3)}{2} = \frac{15}{2} = 7.5\) ✗ (not integer)
• When \(\mathrm{y} = -2\): \(\mathrm{x} = \frac{12 - (-2)}{2} = \frac{14}{2} = 7\) ✓ (integer)
• When \(\mathrm{y} = -1\): \(\mathrm{x} = \frac{12 - (-1)}{2} = \frac{13}{2} = 6.5\) ✗ (not integer)
• When \(\mathrm{y} = 0\): \(\mathrm{x} = \frac{12 - 0}{2} = \frac{12}{2} = 6\) ✓ (integer)
• When \(\mathrm{y} = 1\): \(\mathrm{x} = \frac{12 - 1}{2} = \frac{11}{2} = 5.5\) ✗ (not integer)
• When \(\mathrm{y} = 2\): \(\mathrm{x} = \frac{12 - 2}{2} = \frac{10}{2} = 5\) ✓ (integer)
• When \(\mathrm{y} = 3\): \(\mathrm{x} = \frac{12 - 3}{2} = \frac{9}{2} = 4.5\) ✗ (not integer)
• When \(\mathrm{y} = 4\): \(\mathrm{x} = \frac{12 - 4}{2} = \frac{8}{2} = 4\) ✓ (integer)
• When \(\mathrm{y} = 5\): \(\mathrm{x} = \frac{12 - 5}{2} = \frac{7}{2} = 3.5\) ✗ (not integer)
• When \(\mathrm{y} = 6\): \(\mathrm{x} = \frac{12 - 6}{2} = \frac{6}{2} = 3\) ✓ (integer)
• When \(\mathrm{y} = 7\): \(\mathrm{x} = \frac{12 - 7}{2} = \frac{5}{2} = 2.5\) ✗ (not integer)
• When \(\mathrm{y} = 8\): \(\mathrm{x} = \frac{12 - 8}{2} = \frac{4}{2} = 2\) ✓ (integer)
• When \(\mathrm{y} = 9\): \(\mathrm{x} = \frac{12 - 9}{2} = \frac{3}{2} = 1.5\) ✗ (not integer)
• When \(\mathrm{y} = 10\): \(\mathrm{x} = \frac{12 - 10}{2} = \frac{2}{2} = 1\) ✓ (integer)
• When \(\mathrm{y} = 11\): \(\mathrm{x} = \frac{12 - 11}{2} = \frac{1}{2} = 0.5\) ✗ (not integer)
• When \(\mathrm{y} = 12\): \(\mathrm{x} = \frac{12 - 12}{2} = \frac{0}{2} = 0\) ✓ (integer)

4. Verify integer solutions

Looking at our work above, we can see a clear pattern: x is an integer only when y is even.

Let's list all the valid ordered pairs \(\mathrm{x}, \mathrm{y}\):

• \((12, -12)\)
• \((11, -10)\)
• \((10, -8)\)
• \((9, -6)\)
• \((8, -4)\)
• \((7, -2)\)
• \((6, 0)\)
• \((5, 2)\)
• \((4, 4)\)
• \((3, 6)\)
• \((2, 8)\)
• \((1, 10)\)
• \((0, 12)\)

Counting these pairs, we get 13 ordered pairs.

Process Skill: APPLY CONSTRAINTS - Systematically checking each possibility ensures we don't miss any valid solutions

4. Final Answer

We found exactly 13 ordered pairs \(\mathrm{x}, \mathrm{y}\) where both x and y are integers and satisfy the given system of equations and constraints.

The answer is (D) 13.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the absolute value constraint: Students often misread \(|\mathrm{y}| \leq 12\) as just \(\mathrm{y} \leq 12\), forgetting that absolute value means y can also be negative. This leads them to only consider positive values of y from 0 to 12, missing half of the possible solutions.

2. Not recognizing the integer requirement pattern: Students may attempt to check every single value manually without realizing that \(\mathrm{x} = \frac{12-\mathrm{y}}{2}\) will only be an integer when \((12-\mathrm{y})\) is even, which happens when y is even. Missing this pattern makes the problem much more tedious and error-prone.

3. Confusing the constraint scope: Some students mistakenly think that both x and y must satisfy \(|\mathrm{x}| \leq 12\) and \(|\mathrm{y}| \leq 12\), when the constraint \(|\mathrm{y}| \leq 12\) only applies to y. There is no explicit constraint given for x values.

Errors while executing the approach

1. Arithmetic errors in substitution: When calculating \(\mathrm{x} = \frac{12-\mathrm{y}}{2}\) for negative y values, students frequently make sign errors. For example, when \(\mathrm{y} = -8\), they might calculate \(\mathrm{x} = \frac{12-(-8)}{2} = \frac{12+8}{2}\) incorrectly as \(\frac{12-8}{2} = 2\) instead of the correct answer 10.

2. Miscounting the range of y values: Students often miscount how many integers are in the range from -12 to +12. They might forget to include both endpoints or forget that zero is included, leading to checking the wrong number of y values.

3. Incorrectly identifying integer solutions: When checking if \(\mathrm{x} = \frac{12-\mathrm{y}}{2}\) gives an integer, students may make errors in determining when a fraction is a whole number, especially with negative numbers or when working quickly through many calculations.

Errors while selecting the answer

1. Miscounting valid ordered pairs: After correctly identifying which y values give integer x values, students often miscount when listing the final ordered pairs. They might double-count, skip pairs, or lose track while writing out all 13 solutions.

2. Forgetting the ordered pair requirement: Some students might count the number of valid x values or y values separately instead of counting the ordered pairs \(\mathrm{x},\mathrm{y}\). Since we need both coordinates to specify each solution, this leads to incorrect counting.