document.write( "Question 780814: Joe and Stephanie live 3 miles apart. They both leave their houses at the same time and walk to meet each other. Joe walks at 2.5 miles per hour and Stephanie walks at 3.5 miles per hour. How far will Joe walk before he meets Stephanie?\r
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document.write( "This is my 12 year old sons homework. I am horrible at algebra and don't even know where to start. Please help!!! Thank you in advance to anyone who helps! \n" );
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Algebra.Com's Answer #475691 by MathTherapy(10557) You can put this solution on YOUR website! Joe and Stephanie live 3 miles apart. They both leave their houses at the same time and walk to meet each other. Joe walks at 2.5 miles per hour and Stephanie walks at 3.5 miles per hour. How far will Joe walk before he meets Stephanie?\r \n" ); document.write( "\n" ); document.write( "This is my 12 year old sons homework. I am horrible at algebra and don't even know where to start. Please help!!! Thank you in advance to anyone who helps!\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let distance Joe walked be D \n" ); document.write( "Then distance Stephanie walked = 3 D\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Time equation: \n" ); document.write( "Time Joe took to get to meeting point, equals time it took Stephanie to get to meeting point. Now, since time = distance, divided by speed, or \n" ); document.write( "3.5D = 2.5(3 D) ----- Cross-multiplying \n" ); document.write( "3.5D = 7.5 2.5D \n" ); document.write( "3.5D + 2.5D = 7.5 \n" ); document.write( "6D = 7.5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "D, or distance Joe walked = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You can do the check!! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Further help is available, online or in-person, for a fee, obviously. Send comments, thank-yous, and inquiries to D at MathMadEzy@aol.com \n" ); document.write( " |