Can you help with system of equations problem?

Please forgive me, this may not be 100%. I'm not the best mathematician.

3x + 5y = -34 (multiply by 5)
5x - 3y = 0 (multiply by 3)

15x + 25y = -170 (Subtract bottom line from top)
15x - 9y = 0

So 34y = -170
y = -5

Substitute y = -5 into original and get 3x = -9 so x = -3

So the question now looks like this:
-9 - 25 = -34
-15 + 15 = 0

This is dependant
It is a true statement with infinite solutions eg -34 = -34 or 0 = 0

Edit- Damn lots of people have done it already!

3x + 5y = -34 --------- eq.1
5x - 3y = 0 ---------- eq.2

9x + 15y = – 102 ----- eq.3 (eq. 1 × 3)
25x – 15y = 0 --------- eq. 4 (eq.2 × 5)

34x = – 102
x = – 3

5x - 3y = 0 ---------- eq.2
– 15 – 3y = 0
3y = – 15
y = – 5
Dependent on x and y
--------

3x + 5y = -34 (solve by using elimination)
5x - 3y = 0

3x + 5y = -34
5(3x + 5y) = 5(-34)
15x + 25y = -170

5x - 3y = 0
3(5x - 3y) = 3(0)
15x - 9y = 0

...15x + 25y = -170
‒ 15x - 9y = 0 (subtraction)
-----------------------------------
34y = -170

34y = -170
y = -170/34
y = -5

5x - 3y = 0
5x = 3y
5x = 3(-5)
5x = -15
x = -15/5
x = -3

∴ x = -3 , y = -5
(dependent)

9x + 15y = - 102
25x - 15y = 0-----ADD

34x = - 102
x = - 3

- 9 + 5y = - 34
5y = - 25
y = - 5

x = - 3 , y = - 5
(Dependent)

from 2: 5X = 3Y
X = 3/5Y
substitute in 1: 3 (3/5Y) + 5Y = -34
9/5Y + 5Y = -34
6.8Y = -34
Y = -34/6.8 = -5
back to 2: 5X - 3 * (-5) = 0
5X + 15 =0
5X = -15
X = -15/5 = -3

Check in 1: 3 * (-3) + 5 * (-5) = -9 -25 = -34

3x+5y=-34 ------(1)
5x-3y=0 ----------(2)
from (2) 5x=3y
x=3y/5
put x=3y/5 in (1)

3*3y/5+5y =-34
9y/5+5y=-34
9y+25y/5=-34
34y/5 =-34
34y=-34*5
y= -5

and then x=3*-5/5=-3

3x + 5 = -34 (EQ1)
5x - 3y = 0 (EQ2)

5(EQ1) - 3(EQ2):
15x + 25y = -170
-15x + 9y = 0

34y = -170

y = -5

5x - 3(-5) = 0

5x + 15 = 0

5x = -15

x = -3

Dependent

Check:
3(-3) + 5(-5) = -34

-9 - 25 - 34

-34 = -34
CHECK