tuscany44
Answered

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Need help with this one.
8 +  I  5x  -  3  I  >  10

Sagot :

konrad509
[tex]8 + | 5x - 3 | > 10\\ |5x-3|>2\\ 5x-3>2 \vee 5x-3<-2\\ 5x>5 \vee 5x<1\\ x>1 \vee x<\frac{1}{5}\\ x\in(-\infty,\frac{1}{5})\cup(1,\infty) [/tex]
kayt91

[tex]8 + |5x - 3| > 10[/tex]

[tex]|5x - 3| > 2[/tex]

[tex]|5x| > 5[/tex]

[tex][x] > 1[/tex]


[tex]|5x-3| < -2[/tex]

[tex]|5x| < 1[/tex]

[tex]|x| < \frac{1}{5} [/tex]


[tex]1 > x < \frac{1}{5} [/tex]

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