Solve for the unknown: |2x-5|x-2|| = |x+6| + 3
what a pain
let me see if i can work this out on pencil and paper before writing the solution.
sure. We know it's a pain, even with 4 brains already working on it. Thanks for the help.
i think you have to work in cases. for example the right hand side is a piecewise function, namely \[ |x+6|+3 = \left\{ \begin{array}{lr} x+9 & : x >6\\ -x-3 & : x < 6 \end{array} \right.\]
left hand side is more of a pain
ok i have one of the solutions.
shoot. :)
the left hand side is |2x-5|x-2||
now if x > 2 this is |2x-5(x-2)| = |2x-5x+10| = |-3x+10| yes?
yes.
because of course if x > 2 then |x-2|=x-2
and then?
now |-3x+10| = -3x + 10 if x < 10/3 and |-3x+10|=10-3x if x > 10/3
so if x> 6 both conditions hold and we have the equation \[10-3x=x+9\] solve for x to get \[x=\frac{19}{2}\]
that is if x > 6 then the left hand side is 10-3x and the right hand side is x+9 so solve for x and that is one answer. now we have to do it again for the other side
go on. :)
again on the left if x < 2 then |2x-5|x-2|| is |2x-5(2-x)| is |2x-10+5x| is |7x-10|
hope these steps are clear
i am just replacing |x-2| with (2-x) for x < 2
yes yes. go on
and |7x-10| = 7x-10 if x > 10/7 and |7x-10|=10-7x if x < 10/7
now for x < 10/7 still x > -6 so set \[10-7x=x+9\] \[1=8x\] \[x=\frac{1}{8}\]
and those are the two solutions
what a pain!
btw i made a typo on the first line
yes we got that
*bows* :))) thanks man, that was your second medal :D
i was so impressed with myself for writing a piecewise function that i wrote it incorrectly. it should be \[|x+6|+3 = \left\{ \begin{array}{lr} x+9 & : x >-6\\ -x-3 & : x < -6 \end{array} \right.\]
i had 6 instead of -6
yw
i guess while i am showing off that you should say the line \[7x-10\] for x between 10/7 and 2 does not intersect the line x+9
anyway, we got the main gist of your solution, and was able to work it out. thanks again :-bd
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