SOLUTION: I need help in my algebra class setting up this word problem.
Jake invested $6,000, part at 6% simple interest and part at 4%simple interest for a period of 1 year. How much
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-> SOLUTION: I need help in my algebra class setting up this word problem.
Jake invested $6,000, part at 6% simple interest and part at 4%simple interest for a period of 1 year. How much
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Question 222144: I need help in my algebra class setting up this word problem.
Jake invested $6,000, part at 6% simple interest and part at 4%simple interest for a period of 1 year. How much did she invest at each rate if each account earned the same interest?
You can put this solution on YOUR website! Jake invested $6,000, part at 6% simple interest and part at 4%simple interest for a period of 1 year.
How much did she invest at each rate if each account earned the same interest?
;
Let x = amt invested at 6%
then
(6000-x) = amt invested a 4%
:
int at 6% = int at 4%
.06x = .04(6000-x)
.06x = 240 - .04x
.06x + .04x = 240
.10x = 240
x =
x = $2400 invested at 6%
then
6000 - 2400 = $3600 invested at 4%
;
:
Check solution by find that amt of interest earned by each
.06(2400) = $144
.04(3600) = $144
