3rd
$\displaystyle 9-12x+7 \geq 0 ~\text{and}~9-12x-56<0 $
$\displaystyle 16-12x \geq 0 ~\text{and}~-12x-47<0 $
$\displaystyle 12x \leq 16 ~\text{and}~ 12x > -47 $
$\displaystyle x \leq \frac{4}{3} ~\text{and}~ x > \frac{-47}{12} $
$\displaystyle x \in \left(\frac{-47}{12} , \frac{4}{3} \right] $
Try again for problem number 1.
1. Add 4/9 to each side... * REMEMBER if the DENOMINATOR (# at the bottom) is NOT the same you must find the LCD of both things that you are adding.
SO.... Add (4/9) + (4/81) = ?
2. You should now have the following:
(7/18)t = (product from above)
3. NOW that you have separated your variable from the number divide each side by the reciprocal in front of the variable (so that you can get the variable alone), in this case that would be (18/7)
(18/7)(7/18)t = (product from above)(18/7)
That should give you the answer... if your problem says to simplify into a mixed fraction then do so. I would always just simplify during the multiplication part and just leave my answer as is, even if the numerator was larger than the denominator (improper fraction)
Hope that helps you out!
Number one is 80/63. what you do is when you add 4/9 to 4/81, they must have common denominators, so multiply 4/9 by 9/9 and you get 36/81+4/81 which is 40/81. Not you have 7/18t=40/81. now what you do is divide by 7/18 but in other words it is the same thing if you multiply but the reciprocal on both sides so multiply 18/7 to both sides so on the left you are left with t and on the right you are left to multiply 40/81x18/7. when multiplying fraction, you must multiply numerators only and denominators with denominators and you should get t=720/567 which reduces to 80/63. I hope this helps i assume you know 2 and 3 by the other comments posted so i won't do them. again i hope this helps