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\(2\mathrm{x} + \mathrm{y} = 12\)
\(|\mathrm{y}| \leq 12\)
For how many ordered pairs \(\mathrm{x}, \mathrm{y}\) that are solutions of the system above are x and y both integers?
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
Since \(|\mathrm{y}| \leq 12\), let's think about what values y can take.
The absolute value constraint means:
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).
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:
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 \(\mathrm{x}, \mathrm{y}\):
Counting these pairs, we get 13 ordered pairs.
Process Skill: APPLY CONSTRAINTS - Systematically checking each possibility ensures we don't miss any valid solutions
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.
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.
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.
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.