document.write( "Question 17960: 5(x+3)+9 < or = to 3 (x-2)+6
\n" ); document.write( "5x+15+9 < or = to 3x-6+6
\n" ); document.write( "5x+24 < or = to 3x
\n" ); document.write( "5x/3x+24 < or = to 3x/3x
\n" ); document.write( "this is where I get hung up what do I do with the 5x/3x?
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Algebra.Com's Answer #8667 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Well, you got started ok but your fourth step is in error. You must collect the x-terms (5x & 3x) together on the same side of the = sign.\r
\n" ); document.write( "\n" ); document.write( "\"5x+%2B+24+%3C=+3x\" At this point, you can subtract 24 from both sides.
\n" ); document.write( "\"5x+%3C=+3x+-+24\" Now, subtract 3x from both sides.
\n" ); document.write( "\"2x+%3C=+-24\" Finally, divide both sides by 2.
\n" ); document.write( "\"x+%3C=+-12\"
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