can someone help!!!
- Thread starter kylee
- Start date
this is the answer of your problem, but I believe that you must study in your book
|2s-7|<1
case 1
2s-7 >0
2s>7
s>7/2
there fore
2s-7 < 1
2s<8
s < 4
0 1 2 3 4 5
____l__l__l__l__l__l_____________
l l
the solution are [3.5 , 4]
it finds the other interval
|2s-7|<1
case 1
2s-7 >0
2s>7
s>7/2
there fore
2s-7 < 1
2s<8
s < 4
0 1 2 3 4 5
____l__l__l__l__l__l_____________
l l
the solution are [3.5 , 4]
it finds the other interval
Hello, kylee!
Review your material on Absolute Values.
You don't seem to understand them.
\(\displaystyle \begin{array}{cccc} \text{We have: }& |2x - 7| & < & 1 \end{array}\)
\(\displaystyle \begin{array}{ccccccc} \text{This means:} & -1 & < & 2x-7 & < & 1 \\ \\ \text{Add 7:} & 6 & < & 2x & < & 8 \\ \\ \text{Divide by 2:} & 3 & < & x & < & 4 \end{array}\)
\(\displaystyle \text{The graph looks like this:}\)
. . \(\displaystyle - - - o = = = = o - - -\)
. . . . . . \(\displaystyle 3\qquad\quad\;\;\,4\)
Review your material on Absolute Values.
You don't seem to understand them.
\(\displaystyle \begin{array}{cccc} \text{We have: }& |2x - 7| & < & 1 \end{array}\)
\(\displaystyle \begin{array}{ccccccc} \text{This means:} & -1 & < & 2x-7 & < & 1 \\ \\ \text{Add 7:} & 6 & < & 2x & < & 8 \\ \\ \text{Divide by 2:} & 3 & < & x & < & 4 \end{array}\)
\(\displaystyle \text{The graph looks like this:}\)
. . \(\displaystyle - - - o = = = = o - - -\)
. . . . . . \(\displaystyle 3\qquad\quad\;\;\,4\)
