12(t+6)+8=4(2t-1)+2t 12t+72+8=8t-4+2t 12t-2t-8t=-72-8-4 2t=-84 2t/2=84/2 t=42
3t - 1
2t^2+5t-3=0 (2t-1)(t+3)=0 2t-1=0 and t+3=0 t=.5 and t=-3
6t2+17t+7 = (3t+7)(2t+1) when factored
bcoz one has 3 digits o,n,e and 1 is single so 3+1>2 , thts why 1 plus one is greater than 2
12(t+6)+8=4(2t-1)+2t 12t+72+8=8t-4+2t 12t-2t-8t=-72-8-4 2t=-84 2t/2=84/2 t=42
3t - 1
1 plus
(2t + 1)(3t + 11)
greater than 1
x is greater than 1.
T+(3-2t)=4t+1. Assuming the two T's , T and t are the same variable, lets proceed. t+(3-2t)=4t+1 3-t=4t+1 2-t=4t 2=5t last step is yours
3
2t^2+5t-3=0 (2t-1)(t+3)=0 2t-1=0 and t+3=0 t=.5 and t=-3
You add them but if it equals 1 whole than its not greater than 1😌
This is only true if you accept that 2 is greater than 1...
6t2+17t+7 = (3t+7)(2t+1) when factored