document.write( "Question 521097: 3-4q=10q+10
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Algebra.Com's Answer #346328 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"3-4q=10q%2B10\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"-4q=10q%2B10-3\" Subtract \"3\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"-4q-10q=10-3\" Subtract \"10q\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"-14q=10-3\" Combine like terms on the left side.\r
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\n" ); document.write( "\n" ); document.write( "\"-14q=7\" Combine like terms on the right side.\r
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\n" ); document.write( "\n" ); document.write( "\"q=%287%29%2F%28-14%29\" Divide both sides by \"-14\" to isolate \"q\".\r
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\n" ); document.write( "\n" ); document.write( "\"q=-1%2F2\" Reduce.\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solution is \"q=-1%2F2\" \r
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\n" ); document.write( "\n" ); document.write( "If you need more help, email me at jim_thompson5910@hotmail.com\r
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\n" ); document.write( "\n" ); document.write( "Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you\r
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