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OpenStudy (aaronandyson):
\[3x - 5 \le x + 3 < 5x - 9\]
x belongs to real numbers.
OpenStudy (igreen):
Subtract 3 to all sides
OpenStudy (igreen):
\(3x-5 -3 \le x \le 5x-9-3\)
OpenStudy (igreen):
Simplify, can you do that?
OpenStudy (igreen):
@AaronAndyson
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OpenStudy (aaronandyson):
\[3x - 8 \le x \le 5x - 12\]
OpenStudy (aaronandyson):
@rvc
OpenStudy (aaronandyson):
@ilovewolf
OpenStudy (aaronandyson):
@bibby
@ilovewolf
OpenStudy (aaronandyson):
@dtan5457
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OpenStudy (ilovewolf):
@Directrix
Directrix (directrix):
The thread posts leave off with:
3x -8 ≤ x < 5x - 12 --> the second inequality symbol is < and not ≤ as written above
OpenStudy (aaronandyson):
What does that ? mean
OpenStudy (aaronandyson):
\[3x - 8 \le x \le 5x - 12\]
OpenStudy (aaronandyson):
@Directrix
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Directrix (directrix):
Correct the symbol in your work. See attachment.
OpenStudy (aaronandyson):
ok
so now?
OpenStudy (aaronandyson):
\[3x - 8 \le x \le 5x - 12\]
Directrix (directrix):
In the problem you posted, you wrote "less than"
but now you write less than or equal to.
According to what you posted as the problem.
This is where you should be now:
---------------------------
3x -8 ≤ x < 5x - 12
--------------------------
Directrix (directrix):
3x -8 ≤ x < 5x - 12
means the following:
3x - 8 ≤ x AND x < 5x - 12
3x -x ≤ 8 AND x - 5x < -12
2x ≤ 8 AND -4x < -12
x ≤ 4 AND x > 3
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Directrix (directrix):
Putting these together to get the answer:
x ≤ 4 AND x > 3
becomes
3 < x ≤ 4 ---> Solution to Compound Inequality