its a simple simultaneous equation

Solve the following system:
{3 x-2 y = 7 | (equation 1)
5 x+2 y = 1 | (equation 2)
Swap equation 1 with equation 2:
{5 x+2 y = 1 | (equation 1)
3 x-2 y = 7 | (equation 2)
Subtract 3/5 × (equation 1) from equation 2:
{5 x+2 y = 1 | (equation 1)
0 x-(16 y)/5 = 32/5 | (equation 2)
Multiply equation 2 by 5/16:
{5 x+2 y = 1 | (equation 1)
0 x-y = 2 | (equation 2)
Multiply equation 2 by -1:
{5 x+2 y = 1 | (equation 1)
0 x+y = -2 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{5 x+0 y = 5 | (equation 1)
0 x+y = -2 | (equation 2)
Divide equation 1 by 5:
{x+0 y = 1 | (equation 1)
0 x+y = -2 | (equation 2)
Collect results:
Answer: |x = 1           y = -2

When multiple equations have equal variables.

E₁ = E₂

E₃ = E₄

Then E₁ + E₃ = E₂ + E₄

So...

3x - 2y + 5x + 2y = 7 + 1

8x = 8

x = 1

Substituting x=1 into E1 

3(1) - 2y = 7

3 - 2y = 7

2y = -4

y = -2