Solve the system, using substitution. 3a-b = 11, 2a+3b=0.

3a-b=11 = > b=3a-11

2a+3b=0

2a+3 (3a-11) = 0

2a+9a-33=0

11a=33

=>a=3

=> b=3*3-11=9-11=-2

3a-b=11; 2a+3b=0

Solve 3a-b=11 for b

3a-b+-3a=11+-3a (Add - 3a)

-b=-3a+11

-b - 1 = - 3a+11 / - 1 (Divide - 1)

b=3a-11

Substitute (3a-11) for b in 2a+3b=0

2a+3b=0

2a+3 (3a-11) = 0

11a-33=0 (Simplify)

11a-33+33=0+33 (Add 33)

11a=33

11a/11 = 33 / 11 (Divide by 11)

a=3

Substitute (3) for a in b=3a-11

b=3a-11

b = (3) (3) - 11

b=-2 (Simplify)

b=-2 and a=3