A) To solve this system of linear equations, we can use the substitution method.

First, solve for x in the second equation:
x = 6y - 2

Substitute this expression for x into the first equation:
2(6y - 2) - 3y = 5
12y - 4 - 3y = 5
9y - 4 = 5
9y = 9
y = 1

Now, substitute y = 1 back into x = 6y - 2:
x = 6(1) - 2
x = 6 - 2
x = 4

Therefore, the solution to the system of equations is x = 4 and y = 1.

B) To solve this system of linear equations, we can use the elimination method.

First, multiply the second equation by 2 to make the coefficients of y in both equations equal:
12x - 8y = 22

Now, subtract the first equation from the new second equation:
12x - 8y - (8x + 2y) = 22 - 11
4x - 10y = 11

Now, solve for x:
4x - 10(11) = 11
4x - 10 = 11
4x = 21
x = 21/4 or 5.25

Now, substitute x = 5.25 back into the first equation to solve for y:
8(5.25) + 2y = 11
42 + 2y = 11
2y = -31
y = -31/2 or -15.5

Therefore, the solution to the system of equations is x = 5.25 and y = -15.5.