Algebra 2
posted by Anonymous .
Solve. 4a2≤a+1≤3a+4
4a21≤a+11≤3a+41
4a33≤a≤3a+33
4a/36/3≤a≤3a/3
4a/3(1/a)2≤a≤a(1/a)
4/36≤a≤1
I got lost at this part. I'm not sure what to do now, if i am solving it correctly.
Solve. 3x<45x<5+3x
3x<445x<54+3x
3x/3x<5x<1+3x/3x
I was not sure where to go from here. This is most likely incorrect, the way I am solving it.

4a2 ≤ a+1 ≤ 3a+4
4a2 ≤ a+1 AND a+1 ≤ 3a+4
3a ≤ 3 AND 2a ≤ 3
a ≤ 1 AND a ≥ 3/2
so : 3/2 ≤ a ≤ 1
3x<45x<5+3x
0 < 48x < 5
4 < 8x < 1
1/2 > x > 1/8
or
1/8 < x < 1/2
the reason I did not have to split up the second is that we could subtract 3x from both ends and the x's disappeared.
I had to split up the first, since I could not eliminate the x's from both ends. 
Thank you so much.