solve these 3, two step linear equations using addition methond?

Question

3x-6y=6
9x-3y=8

2/3x-1/2y=3
1/3x-1/4y=3/2

4x+4y=5
2x-8y=-5

Answer 1

1. (1)*-3
-9x+18y=-18
9x-3y=8
adding
15y--10
y=-2/3
sub
x=2/3

2.2/3x-1/2y=3
-2/3x+1/2y=-3/4
no solution

3.4x+4y=5
-4x+16y=10
adding
20y=15
y=3/4
sub
4x+3=5
4x=2
x=1/2

Answer 2

1)
you should multiply the lower eq. by -2 so you get
3x-6y=6
-18x+6y=-16
thus, by addition, of the equations you get
-15x=-10
x=10/15
then substitute x in any initian equation and solve for y and there you go

2) Its the same path, multiply the lower equation by -2
thus, you get
2/3x-1/2y=3
-2/3x+1/2y=-3
this one, when you add the equations, you get
0=0 so you can't solve it 'cause both equations are the same

3) Here, you multiply the first eq. by 2
8x+8y=10
2x-8y=-5
adding them you get..
10x=5
and x=5/10=1/2
substitute x in any initial equation and solve for y

Answer 3

1. 12x-3y=14

2. x-(1/4)y= -3/2

3. 6x-4y=10

Answer 4

Multiply it by way of '-2'. 3x - y = 3 ---> Multiply by way of '-2' -2x + 2y = 6 -6x + 2y = -6 -2x + 2y = 6 ------------------ Subtract -4x = -12 x = 3. replace this 'x' cost in the two of the unique equations to get the 'y' cost. Edit : you won't multiply by way of 3. You 'replace'. 3x - y = 3 ---> replace the cost of 'x'. 3(3) - y = 3 9 - 3 = y ---> y = 6 So answer is 6.(0.33 determination)