document.write( "Question 1159721: Two joggers set out at the same time from their homes 84 miles apart. They agree to meet at a point somewhere in between in four hours. If the rate of one is 3mph faster than the rate of the other.\r
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document.write( "what is the rate of each ? \n" );
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Algebra.Com's Answer #782788 by MathLover1(20850) You can put this solution on YOUR website! \r \n" ); document.write( "\n" ); document.write( "given:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let \n" ); document.write( " \n" ); document.write( "Let \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Equation for slower jogger:\r \n" ); document.write( "\n" ); document.write( "(1) \n" ); document.write( "\n" ); document.write( "Equation for faster jogger:\r \n" ); document.write( "\n" ); document.write( "(2) \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Substitute (1) into (2)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "and\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "then distance that slower jogger runs to meet faster jogger is \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the sum of distances is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |