What is the best approximation of the solution to the system to the nearest valus 2x-3y=-6 and x+2y=8

Given:
2x - 3y = -6  ⇒ identified as 1st equation
x + 2y = 8    ⇒ identified as 2nd equation

Let us first get the value of x using the 2nd equation.
x + 2y = 8
x = 8 - 2y

Substitute x in the 1st equation with its value and find the value of y.
2x - 3y = -6
2(8-2y) - 3y = -6
16 - 4y -3y = -6
-4y - 3y = -6 -16
-7y = - 22
y = -22/-7
y = 22/7 simplified to 3 1/7

Substitute the value of y in the 2nd equation.
x = 8 - 2y
x = 8 - 2(22/7)
x = 8 - 44/7
x = (8 * 7/7) - 44/7
x = 56/7 - 44/7
x = (56-44)/7
x = 12/7 simplified to 1 5/7

To check:  x = 12/7 and y = 22/7  *we need to use the improper fractions.
2x - 3y = -6
2(12/7) - 3(22/7) = -6
24/7 - 66/7 = -6
(24-66)/7 = -6
-42/7 = -6
-6 = -6

x + 2y = 8
12/7 + 2(22/7) = 8
12/7 + 44/7 = 8
(12+44)/7 = 8
56/7 = 8
8 = 8

Ithink if im not mistaking b is -3 but im not a 100% sude