Can anyone please help me with these 2 college algebra questions

~Solve by the Substitution method
3x/2-y/3=-18
3x/4+2y/9=-9


0.1x+0.002y=0.04
5x+y=2

1.) (3x / 2) - (y / 3) = - 18
6(3x / 2) - 6(y / 3) = 6(- 18)
9x - 2y = - 108
9x = 2y - 108
x = 2/9 y - 108/9
x = 2/9 y - 12 ........................... Eq. 1

(3x / 4) + (2y / 9) = - 9
36(3x / 4) + 36(2y / 9) = 36(- 9)
27x + 8y = - 324 ...................... Eq. 2


Sub 2/9 y - 12 from Eq. 1 for x in Eq. 2:

27(2/9 y - 12) + 8y = - 324
6y - 324 + 8y = - 324
14y = 324 - 324
14y = 0
y = 0

and

x = 2/9(0) - 12
x = - 12

Solution Set (- 12, 0)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯


2.) 0.1x + 0.002y = 0.4
0.002y = - 0.1x + 0.4
1000(0.002y) = 1000(- 0.1x) + 1000(0.4)
2y = - 100x + 40
100x + 2y = 40
50x + y = 20 ................. Eq. 1

5x + y = 2
y = - 5x + 2 .................. Eq. 2


Sub - 5x + 2 from Eq. 2 for y in Eq. 1:

50x + (- 5x + 2) = 20
50x - 5x + 2 = 20
45x = 20 - 2
45x = 18
x = 18 / 45
x = 2/5

and

y = - 5(2/5) + 2
y = - 2 + 2
y = 0

Solution Set (2/5, 0)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
 

what college do you go to? I did this in 8th grade haha

substitution means you solve for one variable in terms of the other, then plug that value into the other equation to get a numerical answer.

so for the second one you can see that y=2-5x so you can plug 2-5x in for y in the first equation and solve for it.

i dont have paper so i cant really do the numbers

9x - 2y = -54
27x + 8y = -324

-2y = -9x - 54
y = 4.5x + 27

27x + 9(4.5x + 27) = -324
27x + 40.5x + 243 = -324
67.5x + 243 = -324
67.5x = 567

Divide, solve for x, and ten substitute for y.

y = -5x + 2
0.1x + 0.002(-5x + 2) = 0.04
0.1x - 0.01x + 0.004 = 0.04
0.11x + 0.004 = 0.04
0.11x = 0.036