2 l -x+4 l > 10 solve for X
6l-10>x
theres arent ls haha sorry those are absolute values
those*
oh :D
if you could tell me how to do it that be great i now i have to divide by 2.... but does that change the signn.
yes that is the first step. no it does not change the sign
than you have to divide it to two parts. when x>= 4 than the absolute value does nothing so you get -x+4 >5 -1>x this is contradiction as x cannot be bigger than 4 and smaller than -1 at the same time
whered you get negative one the -x will become positive so i wouldnt have to divide right..? then i would have to do this problem again with flipping the sign and negating the 10
the second part is when 4>=x than the absolute value multiplies it by -1 so you get x-4>5 x>9 this is another contradiction so I must have done something wrong :) time to double check
try this: \[|-x+1|=|1-x|=|x-1|\] now you don't have that annoying minus sign in front of the x. \[2|x-1|>10\] \[|x-1|>5\] \[x-1>5\]or \[x-1<-5\]
there is no contradiction. you have two easy inequalities to solve.
Satellite that answer is incorrect this is a muiltiple choice question and the only answers are with 1 and 9 ... there was no need to move the -x when its in an absolute value sign the - drops anyway which is where the problem may be with your solution
sat it is not -x+1 but -x+4
Hahah thanks Andras that could be the problem as well haha
oops i read it as \[-x+1\] not \[-x+4\] i apologize.
yeah the answer should be x>9 or -1>x
sorry. messed up by not reading carefully.
the choices are 1<x<9 (x<1) or (x>9) (x<-9) or (x>-1) (x<-1)or(x>9)
(x<-1)or(x>9)
ok thanss both for the help :]
thanks*
\[21-x+41>10\] Collect like terms on RHS \[62-x >10\] Subtract 10 from both sides; \[52-x >0\] Add x to both sides \[52>x\] That is;\[x <52\]
Those are absolute values... not 1..
i got the answer already :]..
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