document.write( "Question 93260: hi there,
\n" ); document.write( "I asked the following question before and I had a solution given to me, I do not understand some of it though, some guidance would be great.
\n" ); document.write( "s=u.(v-u)/a + 1/2a((v-u)/a)^2
\n" ); document.write( "it should end up as v^2=u^2+2as \r
\n" ); document.write( "\n" ); document.write( "first step was to multiply by 2a giving....
\n" ); document.write( "2as=2u(v-u)+(v-u)^2\r
\n" ); document.write( "\n" ); document.write( "expanding
\n" ); document.write( "2as= 2uv-2u^2+v^2-2vu+u^2\r
\n" ); document.write( "\n" ); document.write( "apparently by collecting terms this is equal to 2as=v^2-u^2, could someone please explain this last step for me
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Algebra.Com's Answer #67896 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
You were given:
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\n" ); document.write( "\"s=u%28v-u%29%2Fa+%2B+%281%2F2%29%2Aa%2A%28%28v-u%29%2Fa%29%5E2%29\"
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\n" ); document.write( "You need to check the above and make sure this is the problem that you actually meant to
\n" ); document.write( "have explained to you because that is the one that was worked out in the explanation you
\n" ); document.write( "got.
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\n" ); document.write( "First you were told to multiply both sides (all terms) by 2a. The purpose of this multiplication
\n" ); document.write( "was to eventually get rid of the denominators on the right side. It was performed as follows:
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\n" ); document.write( "\"2a%2As+=+2a%2Au%28v-u%29%2Fa+%2B+%28%282a%2A1%29%2F2%29%2Aa%2A%28%28v-u%29%2Fa%29%5E2%29\"
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\n" ); document.write( "On the right side in the second term multiply the \"a\" that is in the middle times the
\n" ); document.write( "numerator of the first factor \"2a%2A1\" which results in \"2a%5E2\" and makes
\n" ); document.write( "the problem
\n" ); document.write( "become:
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\n" ); document.write( "\"2a%2As+=+2a%2Au%28v-u%29%2Fa+%2B+%28%282a%5E2%2A1%29%2F2%29%2A%28%28v-u%29%2Fa%29%5E2%29\"
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\n" ); document.write( "Now in the second term on the right side square out the last factor of \"%28%28v-u%29%2Fa%29%5E2%29\"
\n" ); document.write( "by squaring both its numerator and denominator as follows:
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\n" ); document.write( "You can now substitute the right side of this result in place of \"%28%28v-u%29%2Fa%29%5E2%29\"
\n" ); document.write( "in the problem, and the problem then becomes:
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\n" ); document.write( "At this point you can cancel the factors of 2a in the numerators with corresponding
\n" ); document.write( "factors of 2a in the denominator as follows:
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\n" ); document.write( "and you are left with:
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\n" ); document.write( "\"2a%2As+=+2u%28v-u%29%2Bv%5E2+-+2uv%2Bu%5E2\"
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\n" ); document.write( "Doing the distributed multiplication of the first term on the right side results in:
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\n" ); document.write( "\"2a%2As+=+2uv+-+2u%5E2+%2Bv%5E2+-+2uv%2Bu%5E2\"
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\n" ); document.write( "Notice that on the right side you have terms that can be combined. The 2uv and the -2uv
\n" ); document.write( "cancel each other and therefore, they disappear to leave you with:
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\n" ); document.write( "\"2a%2As+=+-2u%5E2+%2B+v%5E2+%2B+u%5E2\"
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\n" ); document.write( "Then the -2u^2 and the +u^2 combine to result in -u^2 and as a result the equation
\n" ); document.write( "becomes:
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\n" ); document.write( "\"2a%2As+=+v%5E2+-+u%5E2+\"
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\n" ); document.write( "and that's the result you are looking for. Hope this helps.
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