document.write( "Question 261619: Hello! I was hoping you could help me with a question. \"One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading for city A, Going 15 mph faster then the first train. If they meet in 3 hours and 20 minutes how fast were the trains traveling?
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document.write( "I keep trying to put it in the formula 3.2(x)+ 3.2(x-15)=390 from there I get (simplified) 6.4x= 438\r
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document.write( "Please tell me where i'm going wrong
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document.write( "Thank you so much.
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document.write( "Vincent \n" );
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Algebra.Com's Answer #192740 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "3 hours 20 minutes. is not 3.2 but 3 1/3 or 10/3 \n" ); document.write( "10/3*x+10/3(x-15)=390 \n" ); document.write( "10/3(2x-15)=390 \n" ); document.write( "I get x=66 \n" ); document.write( "check \n" ); document.write( "10/3*66+10/3(66-15)=390 \n" ); document.write( "220+10/3*(51)=390 \n" ); document.write( "220+170=390 \n" ); document.write( "ok \n" ); document.write( "check \n" ); document.write( "10/3(2*66-15)=390 \n" ); document.write( "10(2*22-5)=390 \n" ); document.write( "10*(44-5)=390 \n" ); document.write( "10*39=390 \n" ); document.write( "ok \n" ); document.write( " |