How can we solve an equation with more than two absolute values?

By: Guest
Date: Thu-Oct-11-2012-
Response
0
For an expression like:

|2x+4|+|6x+6|=26
we figure out the bounds for |2x+4|
|2x+4|= 2x+4 where 2x+4 >=0, -(2x+4) where 2x+4 < 0
2x+ 4>=0
2x > -4
x >-2
so the expression |2x+4| is
|2x+4|= 2x+4 where x>=-2, -(2x+4) where x < -2

similarly for |6x+6| we get
|6x+6|= 6x+6 where x>=-1, -(6x+6) where x < -1

so we look at our values under our smallest < bound and the work our way up
starting at x <-2, 
we have |2x+4|=-2x-4
|6x+6|=-6x-6 , because if x<-2 then x <-1 must also be true
so we have 
-2x-4 + -6x-6 where x<-2
or simplified
-8x-10 where x <-2
so we now we can solve the equation
-8x-10=26
-8x=36
x=36/-8=-9/2
this is consistent with x<-2, so we accept this as an answer

next smallest bound is x<-1
so we look at numbers between the last < bound and this < bound
i.e. where x>=-2 and x<-1
|2x+4|=2x+4
|6x+6|=-6x-6
so we have
2x+4 + -6x-6 where x>=-2 and x<-1
or simplified
-4x-2 where x>=-2 and x<-1
now we solve
-4x-2=26
-4x=28
x=28/-4=-7
but this is inconsistent with our bounds x>=-2 and x<-1, so we reject this as an answer
As an illustration I will show what happens if we substitute x=-7 into the expression
|2x+4|+|6x+6|=|2(-7)+4|+|6(-7)+6|=|-14+4|+|-42+6|=|-10|+|-36|=10+36=46
Notice that it would have worked if it was -10+36 instead of 10+36

now we try numbers greater than the largest < bound
x >=-1
|2x+4|=2x+4
|6x+6|=6x+6
|2x+4|+|6x+6|=2x+4+6x+6
|2x+4|+|6x+6|=8x+10 where x >-1
we solve this equation now
8x+10=26
8x=16
x=2
x=2 is consistent with x>-1 so we accept this solution

so we have two solutions
x=-9/2 and x=2

substitute the values in to check
|2x+4|+|6x+6|=26
for x=-9/2
|2(-9/2)+4|+|6(-9/2)+6|=|-9+4|+|-27+6|=|-5|+|-21|=5+21=26
for x=2
|2x+4|+|6x+6|=|2(2)+4|+|6(2)+6|=|4+4|+|12+6|=|8|+|18|=26

Hope this makes sense
[d] By: Guest
Date: Thu-Oct-11-2012
Response
What is 1 + 100



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