document.write( "Question 127409: Can someone please help me solve this problem?\r
\n" ); document.write( "\n" ); document.write( "q+9/2 + q-3/3=6\r
\n" ); document.write( "\n" ); document.write( "Thank you.
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Algebra.Com's Answer #93359 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
\"%28%28q%2B9%29%29%2F2\" + \"%28%28q-3%29%29%2F3\" = 6
\n" ); document.write( "We can get rid of the denominators by multiplying the equation by 6:
\n" ); document.write( "6*\"%28%28q%2B9%29%29%2F2\" + 6*\"%28%28q-3%29%29%2F3\" = 6(6)
\n" ); document.write( "Cancel out the denominators and you have:
\n" ); document.write( "3(q+9) + 2(q-3) = 36
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\n" ); document.write( "3q + 27 + 2q - 6 = 36
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\n" ); document.write( "5q + 21 = 36
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\n" ); document.write( "5q = 36 - 21
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\n" ); document.write( "5q = 15
\n" ); document.write( "q = \"15%2F5\"
\n" ); document.write( "q = 3
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\n" ); document.write( "Check in original equation, substitute 3 for q:
\n" ); document.write( "\"%28%283%2B9%29%29%2F2\" + \"%28%283-3%29%29%2F3\" = 6
\n" ); document.write( "\"12%2F2\" + 0 = 6; confirms our solution
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