Deciding When a Linear System Has No Unique Solution
During an algebra review, a student analyzes the linear system 2x + ky = 7 and 4x + 6y = 15. The student substitutes k = 3, notices that the x- and y-coefficients are proportional, and concludes that the system has infinitely many solutions because 'one equation is a multiple of the other.' The tutor points out that the right-hand sides are 7 and 15, not 7 and 14. The student then asks whether some other value of k could produce infinitely many solutions, and whether k = 3 is the only value that fails to give a unique solution. The group is asked to classify the system for all real k, to describe the graphs in the k = 3 case, and to explain how the constant terms control whether proportional coefficient rows mean parallel distinct lines or the same line.
1. Classify the system for all real values of k: unique solution, no solution, or infinitely many solutions. Show the ratio or determinant reasoning.
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The coefficient determinant is (2)(6) − (4)(k) = 12 − 4k. If 12 − 4k ≠ 0, that is k ≠ 3, the system has a unique solution. If k = 3, the system becomes 2x + 3y = 7 and 4x + 6y = 15. Dividing the second equation by 2 gives 2x + 3y = 7.5. The left sides are identical, but 7 ≠ 7.5, so the equations are inconsistent and there is no solution. For infinitely many solutions, the second equation would have to be a nonzero multiple of the first. The x-coefficient ratio forces the multiplier 4/2 = 2, so the y-coefficient would require 6 = 2k, or k = 3, and the constant would require 15 = 2(7) = 14, which is false. Therefore no real k gives infinitely many solutions. The system is inconsistent at k = 3 and has a unique solution for every k ≠ 3.
2. Describe the graphs when k = 3, and explain how the constant terms determine whether proportional coefficient rows mean parallel distinct lines or the same line.
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When k = 3, the equations are 2x + 3y = 7 and 4x + 6y = 15. Solving each for y gives y = (−2/3)x + 7/3 and y = (−2/3)x + 15/6 = (−2/3)x + 5/2. The slopes are equal, so the lines are parallel. The y-intercepts 7/3 and 5/2 are different, so the lines are distinct and do not intersect; hence there is no solution. Proportional coefficient rows guarantee parallel lines. They give the same line only when the constant terms have the same proportionality as the coefficients. Here that would require the second constant to be 2(7) = 14, but it is 15. With 14 instead of 15 and k = 3, the equations would be equivalent and the system would have infinitely many solutions.
3. A student says k = 3 makes the system 'dependent' because the coefficient rows are proportional. Correct the terminology and explain what dependency would require.
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Dependent means the equations represent the same line and the solution set is infinite. At k = 3, the coefficient rows are proportional, but the constant terms are not in the same ratio, so the lines are parallel and distinct. The system is inconsistent, not dependent. For dependency, both the coefficient rows and the constants must be proportional. With the first equation 2x + ky = 7 and the second 4x + 6y = 15, the coefficient ratio 4/2 = 2 forces k = 3 and would also force 15 = 2(7) = 14. Since 15 ≠ 14, dependency is impossible for this system. Proportional coefficients alone indicate parallel or identical lines; the constants decide which.