Complete question is;
Arrange the systems of equations that have a single solution in Increasing order of the x-values in their solutions.
1) 2x + y = 10
x - 3y = -2
2) x + 2y = 5
2x + y = 4
3) x + 3y = 5
6x - y = 11
4) 2x + y = 10
-6x - 3y = -2
Answer:
Arranging the systems in increasing value of x gives;
System 2, system 3, system 1
Step-by-step explanation:
1) 2x + y = 10 - - - (eq 1)
x - 3y = -2 - - - (eq 2)
Multiply eq 1 by 3 to get;
6x + 3y = 30
Add it to eq 2 to get;
7x = 28
x = 28/7
x = 4
2) x + 2y = 5 - - - (eq 1)
2x + y = 4 - - - (eq 2)
Multiply eq 2 by 2 to get;
4x + 2y = 8
Subtract eq 1 from it to get;
3x = 3
x = 3/3
x = 1
3) x + 3y = 5 - - - (eq 1)
6x - y = 11 - - - (eq 2)
Multiply eq 2 by 3 to get;
18x - 3y = 33
Add to eq(1) to get;
19x = 38
x = 38/19
x = 2
4) 2x + y = 10 - - - (eq 1)
-6x - 3y = -2 - - - (eq 2)
Make y the subject in eq 1 to get;
y = 10 - 2x
Put in eq (2);
-6x - 3(10 - 2x) = -2
-6x - 30 + 6x = -2
0 = 0
Thus,no solution exists.
Arranging the systems in increasing value of x gives;
System 2, system 3, system 1
