Answer:
None, each of the four given system of equations have only one solution
Step-by-step explanation:
A system of equations have infinitely many solutions, when the equations represent the same line and therefore, the two equations are the same;
Each of the options are examined as follows;
a. 4·x + 2·y = 5...(1)
-4·x - 2·y = 1...(2)
Expressing both equations with <em>y</em> as the subject, we have;
The value of <em>y</em> in equation (1) is given as follows;
4·x + 2·y = 5
2·y = 5 - 4·x
∴ y = 5/2 - 2·x
The value of <em>y</em> in equation (2) is given as follows;
-4·x - 2·y = 1
-2·y = 1 + 4·x
∴ y = -1/2 - 2·x
Therefore, equation (1) and equation (2) represent the same equation and are therefore equal and have only one solution
b. -10·x + y = 4...(1)
10·x - y = 4...(2)
Therefore, we have;
Adding equation (1) to equation (2) gives;
-10·x + 10·x + y - y = 4 + 4
0 + 0 = 8
0 = 8
Given that 0 ≠ 8, equation (1) is not equal to equation (2) and the two equations have only one solution
c. -8·x + y = 2...(1)
8·x - y = 0...(2)
By examination, the values on the left hand side, LHS, of both equations are the same magnitude but opposite in sign, while the right hand side, RHS, of both equations, have different magnitude
Therefore, the two equations are different and have only one solution
d. -x + 2·y = 6...(1)
7·x - 2·y = 12...(2)
Making <em>y </em>the subject of both equations gives;
Equation (1)
2·y = 6 + x
y = 3 + x/2
Equation (2)
2·y = 12 - 7·x
y = 6 - 3.5·x
Therefore, the two equations are different and have only one solution
