solve each inequality: 3 I 2x-1 I >21
x> 11/104
312x - 11>21 adding 11 to both sides we get 312x> 33 dividing both sides by 312 we get x>33/312 x>11/104 (cancelling out the common factor 3 from both numerator and denominator)
x has to be less than four
and greater than -3
Is that absolute value?
\[3 I 2x-1 I >21 where as 3(2x-11)>-21 or 3(2x-11)>21\]
\[2x-11>-7,2x>-7+11,x>2\]
\[2x-11>7,2x>18,2x \div2>18\div2,x>9\]
ur answers
your using one of the absolute values but not the other it would be 11 instead it would be one
i mean it wouldn't be 11 instead it would be one
okay using these steps solve those
OHHHH!! I read it as 312x-11 !!!! My bad .....
so the final answer would be -3<x<4
x>4 or x<-3
\[3\left| 2x-1 \right|>21\] \[ \left| 2x-1 \right|>7\] \[2x-1>7 or -2x+1>7 \] x>4 Or x<-3
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