document.write( "Question 280654: the sum of two numbers is 53. if three times the smaller is one less than the larger, find the numbers. please show all work im lost \n" ); document.write( "

Algebra.Com's Answer #203983 by oberobic(2304) $\"\"$ $\"About$

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First, lay out what you know or are told.
\n" ); document.write( "x = one number
\n" ); document.write( "y = another number
\n" ); document.write( "x + y = 53 :: given
\n" ); document.write( ".
\n" ); document.write( "Note that later work will be easier if we define y = 53 - x and substitute as needed.
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\n" ); document.write( "x = smaller number :: our decision here is purely arbitrary
\n" ); document.write( "3x = three times the smaller number
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\n" ); document.write( "y = larger number :: logically, it has to be
\n" ); document.write( ".
\n" ); document.write( "3x = y -1 :: 3 times the smaller number is one less than the larger number.
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\n" ); document.write( "substituting for 'y'
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\n" ); document.write( "3x = (53-x) -1
\n" ); document.write( "3x = 53 -x -1
\n" ); document.write( "3x = -x + 52
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\n" ); document.write( "add 'x' to both sides
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\n" ); document.write( "4x = 52
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\n" ); document.write( "divide both sides by 4
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\n" ); document.write( "x = 13
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\n" ); document.write( "substituting once again
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\n" ); document.write( "x + y = 53
\n" ); document.write( "13 + y = 53
\n" ); document.write( "y = 40
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\n" ); document.write( "Answer:
\n" ); document.write( "The two numbers are 13 and 40.
\n" ); document.write( "Their sum is 53
\n" ); document.write( "Three times the smaller number is 3*13 = 39, which is one less than the larger number.
\n" ); document.write( ".
\n" ); document.write( "Done. \n" ); document.write( "

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