Hello, you haven't provided the system of equations, therefore I will show you how to do it for a particular system and you can follow the same procedure for yours.
Answer:
For E1 -> Exactly one
For E2 -> None
For E3 -> Infinitely many
Step-by-step explanation:
Consider the system of equations E1: y = -6x + 8 and 3x + y = 4, replacing equation one in two 3x -6x +8 = 4, solving x = 4/3 and replacing x in equation one y = 0. This system of equations have just one solution -> (4/3, 0)
Consider the system of equations E2: y = -3x + 9 and y = -3x -7, replacing equation one in two -3x + 9 = -3x -7, solving 9 = -3. This system of equations have no solution because the result is a fallacy.
Consider the system of equations E3: 2 = -6x + 4y and -1 = -3x -2y, taking equation one and solving y = 1/2 + 3/2x, replacing equation one in two -1 = -3x -1 +3x, solving -1 = -1. This system of equations have infinitely many solution because we find a true equation when solving .
