document.write( "Question 28520: By what number can each side of 1/3-4x=2/9+7 be mulitplied to clear all fractions? \r
\n" ); document.write( "\n" ); document.write( "A) 6
\n" ); document.write( "B) 3
\n" ); document.write( "C) 12
\n" ); document.write( "D) 9\r
\n" ); document.write( "\n" ); document.write( "When I substitute these numbers in for x the fractions dont seem to clear?
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Algebra.Com's Answer #15521 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Your question seems to indicate some confusion as to the meaning of \"clearing fractions\"\r
\n" ); document.write( "\n" ); document.write( "The clearing of fractions simply means that, after you have multiplied by the number (usually this will be the LCD of the all the fractions in the equation or expression), then you will have only integers as coefficients or contants left in the equation. In your equation:
\n" ); document.write( "\"1%2F3+-+4x+=+2%2F9+%2B+7\" Multiply both sides by the LCD of the two fractions, this will be 9.\r
\n" ); document.write( "\n" ); document.write( "\"9%281%2F3+-+4x%29+=+9%282%2F9+%2B+7%29\" = \"3+-+36x+=+2+%2B+63\"
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