document.write( "Question 1200264: Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was her speed from Tortula to Cactus? \n" ); document.write( "
Algebra.Com's Answer #834437 by MathTherapy(10553)\"\" \"About 
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document.write( "Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was her speed from Tortula to Cactus?\r\n" );
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document.write( "Let speed from Tortula to Cactus, be S\r\n" );
document.write( "Then speed from Cactus to Dry Junction is S + 13\r\n" );
document.write( "Since it took 6 hours to complete the entire journey, we get TIME equation: \"matrix%281%2C3%2C+6%2C+%22=%22%2C+204%2FS+%2B+308%2F%28S+%2B+13%29%29\"\r\n" );
document.write( "                     \"matrix%281%2C3%2C+3%2C+%22=%22%2C+102%2FS+%2B+154%2F%28S+%2B+13%29%29\" ---- Factoring out GCF, 2, in numerators\r\n" );
document.write( "            3S(S + 13) = 102(S + 13) + 154S ----- Multiplying by LCD, S(S + 13) \r\n" );
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document.write( "3S(S - 78) + 17(S - 78) = 0\r\n" );
document.write( "S - 78 = 0             or             3S + 17 = 0____3S = - 17 (ignore)\r\n" );
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document.write( "Speed from Tortula to Cactus, or S = 78 mph
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