Answer:
Therefore the first option gives the correct answer of a system of equations which has exactly one solution.
Explanation:
i) the second option is 2x + 2y = 1 and -2x -2y = 1 . these two equation can be written as 2x + 2y = 1 and 2x + 2y = -1. Therefore we can see that that the expressions for used for the variables x and y are the same but the result is different. This is not possible and so there will be no solutions for these two equations
ii) the third option is 3x + y = -1 and 6x + 2y = -2. therefore these two equations can be written as 3x + y = -1 and 3x + y = -1 (dividing the second equation by 2) and thus we see that both the equations are exactly the same and hence these two equations will have an infinite number of solutions.
iii) the last option is 3x - 2y = 4 and -3x + 2y = 4. these two equations can be written as 3x - 2y = 4 and 3x - 2y = -4. Again we can see in this case that the two equations are exactly the same and therefore these two equations will have an infinite number of solutions.
iv) the remaining option is the first one 2x + y = 3 and 5x - y = 11. If we add the first equation to the second one we 7x = 14 and therefore x = 2. If we substitute x = 2 in the first equation we get (2 × 2) + y = 3, therefore y = -1.
Therefore the first option gives the correct answer of a system of equations which has exactly one solution.