Question 976841: |8x+3|>0
I come up with (-3/8, ∞), but the book has (-∞,-3/8)U(-3/8, ∞).
Can someone walk me through and explain this?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! |8x+3|>0
I come up with (-3/8, ∞), but the book has (-∞,-3/8)U(-3/8, ∞).
Can someone walk me through and explain this?
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(8x+3) inside the absolute sign could be positive or negative and you must solve the equation for both possibilities.
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assume (x+3)>0, just remove the absolute sign and solve the inequality.
8x+3>0
8x>-3
x=>-3/8
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assume (8x+3)<0,affix a negative sign to (8x+3)( and solve the inequality.
-(8x+3)>0
-8x-3>0
-8x>3
divide by (-1) and reverse inequality sign
8x<-3
x<-3/8
number line:
<===)-3/8(====.>
(-∞, -3/8) U (-3/8, ∞)
note: domain does not include -3/8
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