document.write( "Question 172003: divide 184 into two parts such that one third of one part may exceed one seventh of the other part by 8 \n" ); document.write( "

Algebra.Com's Answer #127174 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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divide 184 into two parts such that one third of one part may exceed one seventh of the other part by 8
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\n" ); document.write( "Let one part = x
\n" ); document.write( "and
\n" ); document.write( "the other part = (184-x)
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\n" ); document.write( "\"one third of one part may exceed one seventh of the other part by 8\"
\n" ); document.write( " $\"1%2F3\"$ x = $\"1%2F7\"$ (184-x) + 8
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\n" ); document.write( "Multiply equation by 21 to get rid of the denominators; results:
\n" ); document.write( "7x = 3(184-x) + 21(8)
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\n" ); document.write( "7x = 552 - 3x + 168
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\n" ); document.write( "7x + 3x = 552 + 168
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\n" ); document.write( "10x = 720
\n" ); document.write( "x = $\"720%2F10\"$
\n" ); document.write( "x = 72 is one part
\n" ); document.write( "and
\n" ); document.write( "184 - 72 = 112 is the other part
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\n" ); document.write( ":
\n" ); document.write( "Check these values in the statement
\n" ); document.write( "\"one third of one part may exceed one seventh of the other part by 8\"
\n" ); document.write( " $\"1%2F3\"$ (72) = $\"1%2F7\"$ (112) + 8
\n" ); document.write( " 24 = 16 + 8 \n" ); document.write( "

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