Solve the inequality , find the max possible value for the variable.
3t + 1 < t + 12
2t < 11
t < 5.5
t_max = 5.5
For the second question just solve as two separate inequalities
4 +2 < x <= 7 + 2
6 <x <= 9
x in ]6, 9]
To start off, you should make each fraction have a common denominator.
2*3 = 6
3*2 = 6
The lowest common denominator is six.
so 3 3/6 + 2 2/6 = 5 5/6
Hopefully that helps!
2x-2+x+3+x+11=180
4x+12=180
4x= 168
X= 45