Solving linear systems of equations: addition (elimination method)

6)
multiply 2nd equation by -3

-3x + 15y = -51
 3x + 2y = 0
add
17y = -51
y = -3

 3x + 2y = 0
 3x + 2(-3) = 0
3x = 6
x = 2
answer (2 , -3)

7)
multiply 2nd equation by -2
-2x -4y = -10
 2x + 3y = 6
add
-y = -4
 y = 4

2x + 3y = 6
2x + 3(4) = 6
2x + 12 = 6
2x = -6
x = -3
answer (-3, 4)

8)
multiply 1st equation by 2
6x -2y = 4
x + 2y =3
add
7x = 7
x = 1

x + 2y =3
1 + 2y =3
2y = 2
y = 1
answer (1,1)

9)
multiply 1st equation by 3 :    12x + 15y = 18
multiply 2nd equation by -2:  -12x + 14y = 40
                                            add

29y = 58
y = 2

4x + 5y = 6
4x +5(2) = 6
4x + 10 = 6
4x = -4
x = -1
answer (-1,2)

10)
multiply 2nd equation by 2
32x - 2y = 28
4x + 2y = 8
add
36x = 36
x = 1

4x + 2y = 8
4(1) + 2y = 8
2y = 4
y = 2
answer (1,2)

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.