Solve with addition method?

Question

solve using addition method, indicate whether system is independent, inconsistent, or dependent.
1/3x-1/6y =1/3
1/6x+1/4y=0

Answer 1

I would first simplify each by multiplying by the lowest common denominator.

Multiply the following by 6:
(1/3)x - (1/6)y = 1/3
2x - y = 2

Multiply the following by 12:
(1/6)x + (1/4)y = 0
2x + 3y = 0

To add the equations, you need to make one of the terms the opposite, so I would multiply the top equation by -1:
-2x + y = -2

Now add them:
2x + 3y = 0
-2x + y = -2
----------------
2x -2x + 3y + y = 0 + -2

The x terms cancel out.
4y = -2

Divide both sides by 4:
y = -2/4
y = -1/2

Now solve for x:
2x + 3y = 0
2x + 3(-1/2) = 0
2x -3/2 = 0
2x = 3/2
x = 3/4

Note: If you got to a point where you had something like n = n (e.g. 10 = 10), that would mean any value of x or y would work. This would mean your system is dependent. On the other hand if you got to an untrue statement (e.g. 10 = 3), that would tell you that the system is inconsistent.

Since you got to a solution (x = 3/4, y = -1/2), the system is *independent*.

Answer 2

I think you want to write:

(1/3)x - (1/6)y = 1/3
and
(1/6)x + (1/4)y = 0

If so, multiply the second equation by -2 and add with the first

(1/3)x - (1/6)y = 1/3
(-1/3)x - (1/2)y = 0
-------------------------
-(2/3)y = 1/3 the system has an unique solution

Answer 3

1/3x-1/6y =1/3
1/6x+1/4y=0

1st eq:
(1/3)x=1/3+(1/6)y
x=3(1/3=(1/6)y)
x=1+(1/2)y

Put that in 2nd eq:
(1/6)x+(1/4)y=0
(1/6)*(1+(1/2)y)+(1/4)y=0
1/6 +(1/12)y+(1/4)y=0
(4/12)y=-1/6
y=-(1/6)*(12/4)=-(1/6)*3=-1/2
x=1+(1/2)y=1-(1/2)*(1/2)=3/4

Answer 4

(1)*-1/2
-1/6x+1/12y=-1/6
1/6x+3/12y=0
adding
4/12y=-1/6
1/3y=-1/6
multiplying by 3
y=-1/2
substituting
1/3x+1/12=1/3
multiplying by12
4x+1=4
4x=3
x=3/4