15 points andsolve the system of linear equations. separate the x- and y- values with a coma. 5x+3y=-27 -9x+2y=19

The Answer is x= - 3 , y = -4 

Solution:
Solve the following system:
{5 x + 3 y = -27 | (equation 1)
{2 y - 9 x = 19 | (equation 2)
Swap equation 1 with equation 2:
{-(9 x) + 2 y = 19 | (equation 1)
{5 x + 3 y = -27 | (equation 2)
Add 5/9 × (equation 1) to equation 2:
{-(9 x) + 2 y = 19 | (equation 1)
{0 x+(37 y)/9 = (-148)/9 | (equation 2)

Multiply equation 2 by 9/37:
{-(9 x) + 2 y = 19 | (equation 1)
{0 x+y = -4 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{-(9 x)+0 y = 27 | (equation 1)
{0 x+y = -4 | (equation 2)
Divide equation 1 by -9:
{x+0 y = -3 | (equation 1)
{0 x+y = -4 | (equation 2)
Collect results:
{x = -3                {y = -4

Note the Solve the following system:
{5 x + 3 y = -27 | (equation 1)
{2 y - 9 x = 19 | (equation 2)
Swap equation 1 with equation 2:
{-(9 x) + 2 y = 19 | (equation 1)
{5 x + 3 y = -27 | (equation 2)
Add 5/9 × (equation 1) to equation 2:
{-(9 x) + 2 y = 19 | (equation 1)
{0 x+(37 y)/9 = (-148)/9 | (equation 2)

Multiply equation 2 by 9/37:
{-(9 x) + 2 y = 19 | (equation 1)
{0 x+y = -4 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{-(9 x)+0 y = 27 | (equation 1)
{0 x+y = -4 | (equation 2)
Divide equation 1 by -9:
{x+0 y = -3 | (equation 1)
{0 x+y = -4 | (equation 2)
Collect results:
{x = -3               {y = -4

Please note that the brackets are supposed to span over both eqations, haven found a way to mak it work with this editor. See attachemtn.

5x+3y= -27   | * - 2
-9x+2y= 19   |*3

-10x-6y=54
-27x+6y=57

-37 x=111
x=111/-37
x= -3

5*(-3)+3y= -27
-15+3y=-27
3y=-27+15
3y=-12
y=-12/3
y= -4
Solution : (-3,-4)

Its a right angled triangle not a right triangle