Final answer:
To solve the system of equations with the parameter a: 2x + y = 3 and ax − 3y = 4, we can use the elimination method. The system has a unique solution if 6 - 2a ≠ 0, no solution if 6 - 2a = 0 and 1 ≠ 0, and infinitely many solutions if 6 - 2a = 0 and 1 = 0.
Step-by-step explanation:
To solve the system of equations with the parameter a: 2x + y = 3 and ax − 3y = 4, we can use the method of substitution or elimination. Let's use the elimination method.
Multiply the first equation by 3 and the second equation by 2 to make the coefficients of y in both equations the same:
6x + 3y = 9
2ax − 6y = 8
Now, subtract the second equation from the first equation:
6x + 3y − (2ax − 6y) = 9 - 8
6x + 3y - 2ax + 6y = 1
Combine like terms:
6x - 2ax + 9y = 1
Factor out x from the first two terms and y from the last term:
x(6 - 2a) + 9y = 1
This is a linear equation in the form y = mx + b, where m is the slope.
If the slope is not zero (m ≠ 0), then the system has a unique solution. If the slope is zero (m = 0), then the system has infinitely many solutions.
Therefore, for what values of the parameter a will the system have (i) a unique solution, (ii) no solution, or (iii) infinitely many solutions?
If 6 - 2a ≠ 0, then the system has a unique solution.
If 6 - 2a = 0 and 1 ≠ 0, then the system has no solution.
If 6 - 2a = 0 and 1 = 0, then the system has infinitely many solutions.