document.write( "Question 938084: Divide 46 into two parts such that the sum of the quotients obtained by dividing one part by 7 and the other part by 3may be equal to 10. \n" ); document.write( "

Algebra.Com's Answer #571572 by KMST(5328) $\"\"$ $\"About$

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Let $\"x\"$ and $\"y\"$ be the two parts of $\"46\"$ .
\n" ); document.write( " $\"x%2By=46\"$
\n" ); document.write( "The quotients (if there is no remainder) would be
\n" ); document.write( " $\"x%2F7\"$ and $\"y%2F3\"$ such that
\n" ); document.write( " $\"x%2F7%2By%2F3=10\"$ ---> $\"3x%2B7y=210\"$
\n" ); document.write( "So, we have the system of linear equations
\n" ); document.write( " $\"system%28x%2By=46%2C3x%2B7y=210%29\"$ --> $\"system%28x=46-y%2C3x%2B7y=210%29\"$ --> $\"system%28x=46-y%2C3%2846-y%29%2B7y=210%29\"$ --> $\"system%28x=46-y%2C138-3y%2B7y=210%29\"$ --> $\"system%28x=46-y%2C138%2B4y=210%29\"$ --> $\"system%28x=46-y%2C4y=210-138%29\"$ --> $\"system%28x=46-y%2C4y=72%29\"$ --> $\"system%28x=46-y%2Cy=18%29\"$ --> $\"system%28x=46-18%2Cy=18%29\"$ --> $\"highlight%28system%28x=28%2Cy=18%29%29\"$
\n" ); document.write( "
\n" ); document.write( "NOTE: If the divisions have a quotient and a remainder,
\n" ); document.write( "the problem gets complicated.
\n" ); document.write( "We would have positive integer quotients $\"a\"$ and $\"b\"$
\n" ); document.write( "and positive integer remainders $\"R\"$ and $\"r\"$ .
\n" ); document.write( "The two parts of $\"46\"$ would be
\n" ); document.write( " $\"x=7a%2BR\"$ with $\"1%3C=R%3C=6\"$ , so $\"R\"$ = 1, or 2, or 3, or 4, or 5, or 6, and
\n" ); document.write( " $\"y=3b%2Br\"$ with $\"1%3C=r%3C=2\"$ , so $\"r\"$ = 1, or 2.
\n" ); document.write( "Then $\"a%2Bb=10\"$ and $\"%287a%2BR%29%2B%283b%2Br%29=46\"$ ---> $\"7a%2B3b=46-%28R%2Br%29\"$ .
\n" ); document.write( "We would solve $\"system%28a%2Bb=10%2C7a%2B3b=46-%28R%2Br%29%29\"$ to get
\n" ); document.write( " $\"system%28a=4-%28R%2Br%29%2F4%2Cb=10-a%29\"$ .
\n" ); document.write( "For {{a}} and $\"b\"$ to be integers, with $\"2%3C=R%2Br%3C%2B6%2B2=8\"$ ,
\n" ); document.write( " $\"R%2Br\"$ must be a multiple of $\"4\"$
\n" ); document.write( "If $\"R%2Br=8\"$ it must be
\n" ); document.write( " $\"system%28R=6%2Cr=2%2Ca=4-2=2%2Cb=10-2=8%29\"$ ---> $\"system%28x=7%2A2%2B6%2Cy=3%2A8%2B2%29\"$ ---> $\"system%28x=20%2Cy=26%29\"$
\n" ); document.write( "If $\"R%2Br=4\"$ it could be
\n" ); document.write( " $\"system%28R=2%2Cr=2%2Ca=4-1=3%2Cb=10-3=7%29\"$ ---> $\"system%28x=7%2A3%2B2%2Cy=3%2A7%2B2%29\"$ ---> $\"system%28x=23%2Cy=23%29\"$ , or
\n" ); document.write( " $\"system%28R=3%2Cr=1%2Ca=4-1=3%2Cb=10-3=7%29\"$ ---> $\"system%28x=7%2A3%2B3%2Cy=3%2A7%2B1%29\"$ ---> $\"system%28x=24%2Cy=22%29\"$ \n" ); document.write( "

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