document.write( "Question 73968: 1/(y-4)-2/(y-8)=-1/(y+6) \n" ); document.write( "
Algebra.Com's Answer #53007 by bucky(2189)\"\" \"About 
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\"1%2F%28y-4%29-2%2F%28y-8%29=-1%2F%28y%2B6%29\"
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\n" ); document.write( "Note that we can make everything have a common denominator. The common denominator is
\n" ); document.write( "formed by multiplying together the denominators of all three fractions and is:
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\n" ); document.write( "\"%28y-4%29%2A%28y-8%29%2A%28y%2B6%29\"
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\n" ); document.write( "The first fraction on the left side already has \"%28y-4%29\" as its denominator. Suppose that
\n" ); document.write( "we multiply the first fraction by \"%28%28y-8%29%2A%28y%2B6%29%29%2F%28%28y-8%29%2A%28y%2B6%29%29\". Notice that since this
\n" ); document.write( "multiplier has the same numerator as its denominator it is the same as multiplying by 1.
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\n" ); document.write( "With this multiplication the first fraction becomes:
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\n" ); document.write( "Now let's work on the second fraction on the left side.
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\n" ); document.write( "We do a similar process on the second fraction which is \"-2%2F%28y-8%29\". This fraction
\n" ); document.write( "already has \"y-8\" in the denominator. So this fraction we multiply by
\n" ); document.write( "\"%28%28y-4%29%2A%28y%2B6%29%29%2F%28%28y-4%29%2A%28y%2B6%29%29\"
\n" ); document.write( "because this fraction contains the two factors that are missing for the common denominator.
\n" ); document.write( "The result of this multiplication is:
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\n" ); document.write( "Finally, let's work on the fraction on the right side of the equal sign. This fraction is:
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\n" ); document.write( "\"-1%2F%28y%2B6%29\"
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\n" ); document.write( "and it is missing the two terms \"y-4\" and \"y-8\" if it is to have the common denominator.
\n" ); document.write( "Therefore let's multiply this fraction by \"%28y-4%29%2A%28y-8%29%2F%28%28y-4%29%2A%28y-8%29%29\". The result is:
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\n" ); document.write( "Now we can substitute all 3 of these converted fractions into the original problem to get:
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\n" ); document.write( "Now we can multiply all terms on both sides by the common denominator and that will cancel
\n" ); document.write( "out the common denominator from all terms just leaving us with the numerators as follows:
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\n" ); document.write( "\"%28y-8%29%2A%28y%2B6%29-2%2A%28y-4%29%2A%28y%2B6%29+=+-1%2A%28y-4%29%2A%28y-8%29\"
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\n" ); document.write( "On the right side drop the 1 and just keeping \"-%28y-4%29%2A%28y-8%29\" to get:
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\n" ); document.write( "\"%28y-8%29%2A%28y%2B6%29-2%2A%28y-4%29%2A%28y%2B6%29+=+-%28y-4%29%2A%28y-8%29\"
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\n" ); document.write( "Next do all the multiplications. First, \"%28y-8%29%2A%28y%2B6%29+=+y%5E2+-+2y+-+48\"
\n" ); document.write( "Next,
\n" ); document.write( "\"%28-2%29%2A%28y-4%29%2A%28y%2B6%29+=+%28-2%29%2A%28y%5E2+%2B2y+-24%29=+-2%2Ay%5E2+-4y+%2B+48\"
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\n" ); document.write( "Finally do the multiplication on the right side to get:
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\n" ); document.write( "\"-%28y%5E2+-12y+%2B+32%29+=+-y%5E2+%2B+12y+-+32\"
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\n" ); document.write( "Put all these multiplications into the equation to get:
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\n" ); document.write( "\"y%5E2+-+2y+-+48+-2%2Ay%5E2+-4y+%2B+48+=+-y%5E2+%2B+12y+-+32\"
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\n" ); document.write( "Combine the like terms on the left side. First you have \"y%5E2+-2y%5E2+=+-y%5E2\". Next you have
\n" ); document.write( "\"-2y+-4y+=+-6y\". Finally you have \"-48+%2B48=+0\". When you substitute these results
\n" ); document.write( "on the left side the entire equation becomes:
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\n" ); document.write( "\"-y%5E2+-6y+=+-y%5E2+%2B12y+-32\"
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\n" ); document.write( "This is further reduced by adding \"y%5E2\" to both sides and this eliminates the squared
\n" ); document.write( "terms entirely. What is left is:
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\n" ); document.write( "\"-6y+=+12y+-32\"
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\n" ); document.write( "Subtract 12y from both sides to eliminate the 12y on the right side and the equation becomes:
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\n" ); document.write( "\"-18y+=+-32\"
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\n" ); document.write( "Get rid of the minus signs by multiplying both sides by -1 and you have:
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\n" ); document.write( "\"18y+=+32\"
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\n" ); document.write( "and finally divide both sides by 18 to get:
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\n" ); document.write( "\"y+=+32%2F18\" which reduces to \"y+=+16%2F9\"
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\n" ); document.write( "That's the answer. You can convert it to a proper fraction (1 and 7/9) or a decimal if you
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\n" ); document.write( "It's been a long trip to get to the answer, but there is a lot of good practice involved in
\n" ); document.write( "getting there. Hope this helps you to gain some insight into algebraic manipulations.
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