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How do you do 1/3<1-2x/3<5/4 Am I suppose to find the LCD?
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yes, you can do like this.
Okay, so I got 4<-8x<11 and it has to be in interval notation but the solution says[-11/8,0)
\[\frac{ 1 }{3 }<\frac{ 1-2x }{ 3 }<\frac{ 5 }{4 }\] multiply by 12 4<4-8x<15 4-4<4-8x-4<15-4 0<-8x<11 \[0> x>\frac{ -11 }{8 }~or~\frac{ -11 }{ 8 }<x<0,x \in \left( \frac{ -11 }{8 },0 \right)\]
Oh, okay. I get it. I didn't subtract 4 by itself. Thanks
yw
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