What about it? Are we solving?
The first:
3b=2+a ----> a = 3b - 2
The second:
3a-b=2+a+2b
2a - 3b = 2
now sub in the first,
2(3b-2) - 3b = 2
6b - 4 - 3b = 2
3b = 6
b = 2, then in a = 3b-2 ,
a = 4
3b=2+a
3a-b=2+a+2b
2 answers
First equation:
3 b = 2 + a
Subtract 2 to both sides
3 b - 2 = a
a = 3 b - 2
Second equation:
3 a - b = 2 + a + 2 b
Subtract a to both sides
2 a - b = 2 + 2 b
Add b to both sides
2 a = 2 + 3 b
2 a = 3 b + 2
Since: a = 3 b - 2
replace a with 3 b - 2 in this equation
2 ( 3 b - 2 ) = 3 b + 2
6 b - 4 = 3 b + 2
Subtract 3 b to both sides
3 b - 4 = 2
Add 4 to both sides
3 b = 6
b = 6 / 3
b = 2
Put this value in:
a = 3 b - 2
a = 3 • 2 - 2
a = 6 - 2
a = 4
3 b = 2 + a
Subtract 2 to both sides
3 b - 2 = a
a = 3 b - 2
Second equation:
3 a - b = 2 + a + 2 b
Subtract a to both sides
2 a - b = 2 + 2 b
Add b to both sides
2 a = 2 + 3 b
2 a = 3 b + 2
Since: a = 3 b - 2
replace a with 3 b - 2 in this equation
2 ( 3 b - 2 ) = 3 b + 2
6 b - 4 = 3 b + 2
Subtract 3 b to both sides
3 b - 4 = 2
Add 4 to both sides
3 b = 6
b = 6 / 3
b = 2
Put this value in:
a = 3 b - 2
a = 3 • 2 - 2
a = 6 - 2
a = 4