document.write( "Question 300921: Solve for a (steps would be appreciated)\r
\n" ); document.write( "\n" ); document.write( "\"%281%29%2F%28a%29%2B%281%29%2F%28b%29=+%281%29%2F%28c%29\"
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Algebra.Com's Answer #215825 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
\"%281%29%2F%28a%29%2B%281%29%2F%28b%29=+%281%29%2F%28c%29\"
\n" ); document.write( "Multiply by \"abc\".
\n" ); document.write( "\"bc%2Bac=ab\"
\n" ); document.write( "Move \"a\" terms on one side, everything else on the other.
\n" ); document.write( "\"ac-ab=-bc\"
\n" ); document.write( "Factor out the \"a\".
\n" ); document.write( "\"a%28c-b%29=-bc\"
\n" ); document.write( "\"a=-bc%2F%28c-b%29\"
\n" ); document.write( "\"a=bc%2F%28b-c%29\"
\n" ); document.write( "Assumption is that \"b-c\" doesn't equal zero.
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