Question

|8t-9| less than or equal to 6
Find the solution and answer in interval notation.
Please explain how to do it, I'm really confused!

Answer 1

\[|8t-9| \le 6\]

Answer 2

\[ \large |a|\leq b\Leftrightarrow\begin{cases}
b\geq0\\\text{and}\\ -b\leq a\leq b
\end{cases} \]

Answer 3

Can you explain? I haven't seen that before

Answer 4

\[ \large |8t-9|\leq6 \]
becomes
\[ \large -6\leq 8t-9\leq 6 \]

Answer 5

Alrighty, and then what do you do? o:

Answer 6

first add 9 to everything
then divide by 8

Answer 7

Oh okay I got it! Thank you!

Answer 8

u r welcome

Answer 9

@helder_edwin the question has mod in it, dont you have to square it to remove the mod
I dont know how you did it but please tell me what you thin of this method??
\[\left| 8t-9 \right|\le6\]\[(8t-9)^{2}\le(6)^{2}\]\[(64t^{2}-144t+81)\le36\]\[64t^{2}-144t+45\le0\]I am not sure what to do next, but shouldn't it start like this??

Answer 10

@Anas.P

Answer 11

@Anas.P r u there?

Answer 12

well. what u did u fine.
but see what Insist did:
\[ \large |8t-9|\leq6 \]
\[ \large -6\leq 8t-9\leq6 \]
\[ \large 3\leq 8t\leq15 \]
\[ \large \frac{3}{8}\leq t\leq\frac{15}{8} \]
so the solution is \([3/8,15/8]\).

Answer 13

whereas what u did is incomplete. u haven't solved the inequation just yet.

Answer 14

the way Insist solved the inequation is easier.

Answer 15

to finish what u started:
now factor the quadratic polynomial u got:
\[ \large 64t^2-144t+45\leq0 \]
\[ \large (8x-15)(8x-3)\leq0 \]
from this u get the points
\[ \large x=3/8\qquad\vee\qquad x=15/8 \]

Answer 16

hey thanks.... gotya...
nice way of solving, Thanks a lot..