document.write( "Question 46252: Problem 25 I got wrong.Thank You!!\r
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Algebra.Com's Answer #30692 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Simplify:
\n" ); document.write( "\"%285-%28a%2Fb%29%29%2F%2815%2Fb%29\"
\n" ); document.write( "Simplify the numerator first:\r
\n" ); document.write( "\n" ); document.write( "\"5-%28a%2Fb%29+=+%285b-a%29%2Fb\" You get this by putting the fractions over a common denominator, then subtracting. The common denominator is b. The second fraction is already over b so don't change it.
\n" ); document.write( "To get the first fraction over b you multiply the 5 by \"b%2Fb\" to get \"5b%2Fb\". Now you have both parts over the common denominator of b and you can subtract.
\n" ); document.write( "\"5-%28a%2Fb%29+=+%285b-a%29%2Fb\" \r
\n" ); document.write( "\n" ); document.write( "So now you have:
\n" ); document.write( "\"%28%285b-a%29%2Fb%29%2F%2815%2Fb%29\"\r
\n" ); document.write( "\n" ); document.write( "Remember, to divide two fractions, you copy the first, flip the sign from divide to multiply, and flip (invert) the second fraction. So it now becomes:
\n" ); document.write( "\"%28%285b-a%29%2Fb%29%2A%28b%2F15%29\" Cancel the b's
\n" ); document.write( "\"%285b-a%29%2F15\"...and this is as far as you can go!
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