Find the sum of all positive integral values of a for which every solution to the system of equation x + ay = 3 and ax + 4y = 6 satisfy the inequalities x > 1, y > 0.
Step-by-Step Solution
Key Concept: Solve the system of linear equations using Cramer's rule or substitution to express x and y in terms of a, then apply the given inequalities x > 1 and y > 0 to find the valid range for the positive integer a.
The system is x + ay = 3 and ax + 4y = 6. The determinant of the coefficient matrix is D = 4 - a^2. For a unique solution, D != 0, so a != 2 and a != -2. Using Cramer's rule: x = (12 - 6a) / (4 - a^2) = 6(2 - a) / (2 - a)(2 + a) = 6 / (2 + a). y = (6 - 3a) / (4 - a^2) = 3(2 - a) / (2 - a)(2 + a) = 3 / (2 + a). Given x > 1, 6 / (2 + a) > 1 => 6 > 2 + a => a < 4. Given y > 0, 3 / (2 + a) > 0 => 2 + a > 0 => a > -2. Since a is a positive integer, a can be 1, 2, 3. However, a != 2 for a unique solution. Thus, a can be 1 or 3. The sum is 1 + 3 = 4.
Correct Answer: 4