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