Question 145651: Help with 2 word problems
Solve the following applications problems using any method
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?
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?
With both of these I can get so far and I get stuck Thanks
Answer by Earlsdon(6294)   (Show Source):
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.
The interest earned on these two amounts can be expressed as, after changing the percentages to their decimal equivalents:
0.09x is the interest earned at 9%.
0.1($15,000-x) is the interest earned at 10%
The sum of these two amounts is $1,432,so we can set up an equation to find x:
0.09x+0.1($15,000-x) = $1,432 Simplify.
0.09x+1500-0.1x = 1432 Combine like-terms.
(0.09x-0.1x)+1500 = 1432
-0.01x+1500 = 1432 Subtract 1500 from both sides.
-0.01x = -68 Divide both sides by -0.01
x = 6800, so...
$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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2) Use the distance formula: d = rt where d = distance traveled, r = rate of speed, and t = time of travel.
Car 1:
 and...
Car 2:
For car 1, substitute  and for car 2 substitute  and 
The time of travel,t, is the same for both cars, so...
1) 
2)  , so we can write...
 Simplify.
 Add 1500 to both sides.
 Subtract  from both sides.
 Divide both sides by 25.
The first car was traveling at 60mph and the second car was traveling at (60-5) = 55mph.
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