Can you help me with this problem. I keep getting no solution. It's a simple problem.
3(b-1)<3b+3
Please help!
3b-3 < 3b+3
Logically, this means that any number b will satisfy this equation.
3(b-1) < 3b + 3
3b - 3 < 3b + 3
3b - 3b < 3 + 3
0(b) < 6 => There are infinitely many solutions, since any number multiplied with 0 gives 0, which is less than 6.
3(b-1) < 3b+3
-3 < 3
Infinitely many solutions
Simplify , 3b-3<3b+3 subtract 3b from each side,
=> -3=+3
This is impossible so there is no solution, as you suggest
3b-3<3b+3
0<6
which is true. so any value of b works.
Comments
3b-3 < 3b+3
Logically, this means that any number b will satisfy this equation.
3(b-1) < 3b + 3
3b - 3 < 3b + 3
3b - 3b < 3 + 3
0(b) < 6 => There are infinitely many solutions, since any number multiplied with 0 gives 0, which is less than 6.
3(b-1) < 3b+3
3b-3 < 3b+3
-3 < 3
Infinitely many solutions
Simplify , 3b-3<3b+3 subtract 3b from each side,
=> -3=+3
This is impossible so there is no solution, as you suggest
3b-3<3b+3
0<6
which is true. so any value of b works.