abs(3x+1)+abs(2x-3)=11=3x+1-3x+10 so
abs(3x+1)/(3x+1)+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1) or
pm1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)
Now we have two options:
a)
1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)->abs(2x-3)/(3x+1)=(10-3x)/(3x+1)
b)
-1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)->abs(2x-3)/(3x+1)=2+(10-3x)/(3x+1)
Following with a) we have
a-1)
abs(2x-3)=10-3x=2x-3-5x+13 or
abs(2x-3)/(2x-3)=1+(13-5x)/(2x-3) or
pm1=1+(13-5x)/(2x-3) with two options:
a-1-1)
1=1+(13-5x)/(2x-3)->13-5x=0->color(red)(x=13/5)
a-1-2)
-1=1+(13-5x)/(2x-3)->0=2+(13-5x)/(2x-3)->color(red)(x=7)
Now following with b)
abs(2x-3)/(3x+1)=2+(10-3x)/(3x+1)->abs(2x-3)=6x+2+10-3x so
abs(2x-3)=3x+12 then
b-1)
abs(2x-3)=2x-3+x+15->abs(2x-3)/(2x-3)=1+(x+15)/(2x-3) or
pm1=1+(x+15)/(2x-3) with two options
b-1-1)
1=1+(x+15)/(2x-3)->color(red)(x=-15)
b-1-2)
-1=1+(x+15)/(2x-3)->0=2+(x+15)/(2x-3)->color(red)(x=-9/5)
After checking the found values, we pick the feasible values x = -9/5 and x = 13/5
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