Questions on Word Problems: Evaluation, Substitution answered by real tutors!

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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 : 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(3748) About Me  (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:
d[1] = r[1]*t and...
Car 2:
d[2] = r[2]*t
For car 1, substitute d[1] = 300 and for car 2 substitute r[2] = r[1]-5 and d[2] = 275
The time of travel,t, is the same for both cars, so...
1) t = 300/r[1]
2) t = 275/(r[1]-5), so we can write...
300/r[1] = 275/(r[1]-5) Simplify.
300(r[1]-5) = 275(r[1])
300r[1]-1500 = 275r[1] Add 1500 to both sides.
300r[1] = 275r[1]+1500 Subtract 275r[1] from both sides.
25r[1] = 1500 Divide both sides by 25.
r[1] = 60
The first car was traveling at 60mph and the second car was traveling at (60-5) = 55mph.