HELP PLEAAAAASE! how do i solve these

Q1)

7/9x+ y= 53/9
-1/8x- 1/4y= -7/4

Q2)

2x+2y-4z=6
y+5z=-35
3x-2y=-2

Thanks!

(7/9)x+y = 53/9
9[(7/9)x+y] = 9(53/9)
7x+9y = 53
9y = 53-7x
y = (53-7x)/9

(-1/8)x-(1/4)y = -7/4
-8[(-1/8)x-(1/4)y] = -8(-7/4)
x+2y = 14
2y = 14-x
y = (14-x)/2

(53-7x)/9 = (14-x)/2
2(53-7x) = 9(14-x)
106-14x = 126-9x
106 = 126-9x+14x
106 = 126+5x
106-126 = 5x
-20 = 5x
-20/5 = x
-4 = x

(7/9)x+y = 53/9
(7/9)(-4)+y = 53/9
(-28/9)+y = 53/9
y = (53/9)-(-28/9)
y = (53/9)+(28/9)
y = 81/9
y = 9

Answer: x=-4 and y=9
______________________________________…

2x+2y-4z = 6
2(x+y-2z) = 6
x+y-2z = 6/2
x+y-2z = 3
y = 3-x+2z

3x-2y = -2
3x = -2+2y
x = (-2+2y)/3
x = (2y-2)/3

y = 3-x+2z
y = 3-[(2y-2)/3]+2z
y = (9/3)-[(2y-2)/3]+(6z/3)
y = [9-(2y-2)+6z]/3
y = (9-2y+2+6z)/3
y = (11-2y+6z)/3
3y = 11-2y+6z
3y+2y = 11+6z
5y = 11+6z
y = (11+6z)/5

y+5z = -35
[(11+6z)/5]+5z = -35
[(11+6z)/5]+(25z/5) = -35
(11+6z+25z)/5 = -35
(11+31z)/5 = -35
11+31z = 5(-35)
11+31z = -175
31z = -175-11
31z = -186
z = -186/31
z = -6

y+5z = -35
y+5(-6) = -35
y-30 = -35
y = -35+30
y = -5

3x-2y = -2
3x-2(-5) = -2
3x+10 = -2
3x = -2-10
3x = -12
x = -12/3
x = -4

Answer: x=-4, y=-5, and z=-6

Try the elimination method. First start out by finding the two numbers that one is positive and one is negative with the same variable. In this case it is x. Now focus on making these two numbers the same by multiplying the whole second (or the number thats negative) equation so that the two numbers attached to x are the opposite: Example 7/9 and -7/9. And remember to time the whole equation by this number. Once you find that number and apply it to your equation, remove any like numbers and take it from there by combining equations. As for combining a single y, remember any variable without a number is always equal to one. Once you have them combined, divide the number attached to the y by the number on the other side of the equation. You then should have your answer. As for question 2... I'm not sure.