document.write( "Question 145651: Help with 2 word problems
\n" ); document.write( "Solve the following applications problems using any method
\n" ); document.write( " A couple working towards retirement made two investments totaling $15,000. In one year, these investments yielded $1432 in simple interest. Part of the money was invested at 9% and the rest at 10% How much was invested at each rate?\r
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\n" ); document.write( "\n" ); document.write( "A car travels 300 miles in the same amount of time that a car traveling 5 miles and hour slower travels 275 miles. At what speed (rate) is each car traveling?
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\n" ); document.write( "\n" ); document.write( "With both of these I can get so far and I get stuck Thanks
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Algebra.Com's Answer #106286 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
1) Let x = the amount invested at 9% simple interest. Then $15,000-x = the amount invested at 10% simple interest.
\n" ); document.write( "The interest earned on these two amounts can be expressed as, after changing the percentages to their decimal equivalents:
\n" ); document.write( "0.09x is the interest earned at 9%.
\n" ); document.write( "0.1($15,000-x) is the interest earned at 10%
\n" ); document.write( "The sum of these two amounts is $1,432,so we can set up an equation to find x:
\n" ); document.write( "0.09x+0.1($15,000-x) = $1,432 Simplify.
\n" ); document.write( "0.09x+1500-0.1x = 1432 Combine like-terms.
\n" ); document.write( "(0.09x-0.1x)+1500 = 1432
\n" ); document.write( "-0.01x+1500 = 1432 Subtract 1500 from both sides.
\n" ); document.write( "-0.01x = -68 Divide both sides by -0.01
\n" ); document.write( "x = 6800, so...
\n" ); document.write( "$6,800 was invested at 9% simple interest and $15,000-$6,800 = $8,200 was invested at 10% simple interest.
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\n" ); document.write( "2) Use the distance formula: d = rt where d = distance traveled, r = rate of speed, and t = time of travel.
\n" ); document.write( "Car 1:
\n" ); document.write( "\"d%5B1%5D+=+r%5B1%5D%2At\" and...
\n" ); document.write( "Car 2:
\n" ); document.write( "\"d%5B2%5D+=+r%5B2%5D%2At\"
\n" ); document.write( "For car 1, substitute \"d%5B1%5D+=+300\" and for car 2 substitute \"r%5B2%5D+=+r%5B1%5D-5\" and \"d%5B2%5D+=+275\"
\n" ); document.write( "The time of travel,t, is the same for both cars, so...
\n" ); document.write( "1) \"t+=+300%2Fr%5B1%5D\"
\n" ); document.write( "2) \"t+=+275%2F%28r%5B1%5D-5%29\", so we can write...
\n" ); document.write( "\"300%2Fr%5B1%5D+=+275%2F%28r%5B1%5D-5%29\" Simplify.
\n" ); document.write( "\"300%28r%5B1%5D-5%29+=+275%28r%5B1%5D%29\"
\n" ); document.write( "\"300r%5B1%5D-1500+=+275r%5B1%5D\" Add 1500 to both sides.
\n" ); document.write( "\"300r%5B1%5D+=+275r%5B1%5D%2B1500\" Subtract \"275r%5B1%5D\" from both sides.
\n" ); document.write( "\"25r%5B1%5D+=+1500\" Divide both sides by 25.
\n" ); document.write( "\"r%5B1%5D+=+60\"
\n" ); document.write( "The first car was traveling at 60mph and the second car was traveling at (60-5) = 55mph.\r
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