document.write( "Question 51422: One number is 5 more than another. Five times the smaller is 2 more
\n" ); document.write( "than twice the larger. Find the numbers.\r
\n" ); document.write( "\n" ); document.write( "Bill has twice as much money as bob. Paul has $12 more than Bill.
\n" ); document.write( "Together they have $92. How much money does Bob have?\r
\n" ); document.write( "\n" ); document.write( "Find two consecutive whole numbers that total 93.\r
\n" ); document.write( "\n" ); document.write( "WORK DONE :\r
\n" ); document.write( "\n" ); document.write( "NOTHING IM COMPLETELY LOST!!!!!!! \r
\n" ); document.write( "\n" ); document.write( "I do not remmeber what to do. \r
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Algebra.Com's Answer #34359 by rchill(405) $\"\"$ $\"About$

You can put this solution on YOUR website!
You have three separate problems here -- you should probably ask one problem at a time in the future.
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\n" ); document.write( "In the first problem we're told that one number is 5 more than another. Let $\"x\"$ represent the number, then $\"x%2B5\"$ represents the other. Five times the smaller (i.e., $\"5x\"$ ) is 2 more than twice the larger (i.e., $\"2%28x%2B5%29%2B2\"$ ). This produces the equation $\"5x=2%28x%2B5%29%2B2\"$ . Now we just solve for $\"x\"$ . Expanding the equation produces $\"5x=2x%2B10%2B2\"$ which simplifies to $\"5x=2x%2B12\"$ . Subtracting $\"2x\"$ from both sides produces $\"3x=12\"$ . Now divide both sides by $\"3\"$ yields $\"x=4\"$ . This means the smaller number is 4 and the other number is 5 more than that; i.e. 4+5, or 9.
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\n" ); document.write( "For the second problem, let $\"x\"$ represent the money for Bob. Bill has twice that amount, or $\"2x\"$ . Paul has $12 more than Bill, or $\"2x%2B12\"$ . Together they have $92, represented by the equation $\"x%2B2x%2B2x%2B12=92\"$ . We combine like terms to get $\"5x%2B12=92\"$ . Subtracting 12 from both sides yields $\"5x=80\"$ . Now divide both sides by 5 to produce $\"x=16\"$ . That means Bob has $16. Bill has twice that, or $32; Paul has $12 more than Bill, or $44. Adding $16+$32+$44 produces the sum of $92.
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\n" ); document.write( "For the third problem, let $\"x\"$ represent one of the numbers. Because the numbers are consecutive, the other number is represented by $\"x%2B1\"$ . Their sum must total 93, represented by the equation $\"x%2Bx%2B1=93\"$ . Combining like terms produces $\"2x%2B1=93\"$ . Solving for $\"x\"$ as in the process above produces $\"x=46\"$ . That means the second number is one more than that, or 47. Sure enough, $\"46%2B47=93\"$ . \n" ); document.write( "

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