document.write( "Question 976841: |8x+3|>0\r
\n" ); document.write( "\n" ); document.write( "I come up with (-3/8, ∞), but the book has (-∞,-3/8)U(-3/8, ∞).\r
\n" ); document.write( "\n" ); document.write( "Can someone walk me through and explain this? \n" ); document.write( "
Algebra.Com's Answer #598340 by lwsshak3(11628)\"\" \"About 
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|8x+3|>0
\n" ); document.write( "I come up with (-3/8, ∞), but the book has (-∞,-3/8)U(-3/8, ∞).
\n" ); document.write( "Can someone walk me through and explain this?
\n" ); document.write( "***
\n" ); document.write( "(8x+3) inside the absolute sign could be positive or negative and you must solve the equation for both possibilities.
\n" ); document.write( "..
\n" ); document.write( "assume (x+3)>0, just remove the absolute sign and solve the inequality.
\n" ); document.write( "8x+3>0
\n" ); document.write( "8x>-3
\n" ); document.write( "x=>-3/8
\n" ); document.write( "..
\n" ); document.write( "assume (8x+3)<0,affix a negative sign to (8x+3)( and solve the inequality.
\n" ); document.write( "-(8x+3)>0
\n" ); document.write( "-8x-3>0
\n" ); document.write( "-8x>3
\n" ); document.write( "divide by (-1) and reverse inequality sign
\n" ); document.write( "8x<-3
\n" ); document.write( "x<-3/8
\n" ); document.write( "number line:
\n" ); document.write( "<===)-3/8(====.>
\n" ); document.write( "(-∞, -3/8) U (-3/8, ∞)
\n" ); document.write( "note: domain does not include -3/8 \n" ); document.write( "
\n" );