document.write( "Question 199563: 3. Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle. \r
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Algebra.Com's Answer #149972 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Remember, the distance-rate-time formula is \r
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\n" ); document.write( "\n" ); document.write( "\"d=rt\"\r
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\n" ); document.write( "\n" ); document.write( "Since he \"traveled 150 miles at a certain speed\", this means that the first equation is \"150=rt\" (simply plug in d=150)\r
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\n" ); document.write( "\n" ); document.write( "The statement \"Had gone 20mph faster, the trip would have taken 2 hour less\", tells us that the next equation is \"150=%28r%2B20%29%28t-2%29\" (the distance is the same, but the speed is now 20 mph faster and the time is 2 hours shorter)\r
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\n" ); document.write( "\n" ); document.write( "The goal is to use this system to solve for \"r\" (and if we want, \"t\" also)\r
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\n" ); document.write( "\n" ); document.write( "\"150=rt\" Start with the first equation.\r
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\n" ); document.write( "\n" ); document.write( "\"150%2Fr=t\" Divide both sides by r.\r
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\n" ); document.write( "\n" ); document.write( "\"t=150%2Fr\" Rearrange the equation\r
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\n" ); document.write( "\n" ); document.write( "\"150=%28r%2B20%29%28t-2%29\" Move onto the second equation.\r
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\n" ); document.write( "\n" ); document.write( "\"150=%28r%2B20%29%28150%2Fr-2%29\" Plug in \"t=150%2Fr\" \r
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\n" ); document.write( "\n" ); document.write( "\"150=150-2r%2B3000%2Fr-40\" FOIL\r
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\n" ); document.write( "\n" ); document.write( "\"150r=150r-2r%5E2%2B3000-40r\" Multiply EVERY term by the LCD \"r\" to clear out the fraction.\r
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\n" ); document.write( "\n" ); document.write( "\"0=150r-2r%5E2%2B3000-40r-150r\" Subtract \"600r\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"0=-2r%5E2-40r%2B3000\" Combine and rearrange the terms.\r
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\n" ); document.write( "\n" ); document.write( "Notice we have a quadratic in the form of \"Ar%5E2%2BBr%2BC\" where \"A=-2\", \"B=-40\", and \"C=3000\"\r
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\n" ); document.write( "\n" ); document.write( "Let's use the quadratic formula to solve for \"r\":\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29\" Start with the quadratic formula\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-%28-40%29+%2B-+sqrt%28+%28-40%29%5E2-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29\" Plug in \"A=-2\", \"B=-40\", and \"C=3000\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+%28-40%29%5E2-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29\" Negate \"-40\" to get \"40\". \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+1600-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29\" Square \"-40\" to get \"1600\". \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+1600--24000+%29%29%2F%282%28-2%29%29\" Multiply \"4%28-2%29%283000%29\" to get \"-24000\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+1600%2B24000+%29%29%2F%282%28-2%29%29\" Rewrite \"sqrt%281600--24000%29\" as \"sqrt%281600%2B24000%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+25600+%29%29%2F%282%28-2%29%29\" Add \"1600\" to \"24000\" to get \"25600\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+sqrt%28+25600+%29%29%2F%28-4%29\" Multiply \"2\" and \"-2\" to get \"-4\". \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B-+160%29%2F%28-4%29\" Take the square root of \"25600\" to get \"160\". \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%2840+%2B+160%29%2F%28-4%29\" or \"r+=+%2840+-+160%29%2F%28-4%29\" Break up the expression. \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28200%29%2F%28-4%29\" or \"r+=++%28-120%29%2F%28-4%29\" Combine like terms. \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+-50\" or \"r+=+30\" Simplify. \r
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\n" ); document.write( "\n" ); document.write( "So the possible solutions are \"r+=+-50\" or \"r+=+30\" \r
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\n" ); document.write( "\n" ); document.write( "Since a negative speed doesn't make any sense, this means that we must ignore \"r+=+-50\"\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the only solution is \"r+=+30\" which means that he was traveling 30 mph. So you are correct, good job.
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