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How much was invested in each account?

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3 Answers

x=amount invested at 14%
(18,000-x)= amount invested at 15%
 
0.14x=(18,000-x)(0.15)+1357.10
0.14x=2700-0.15x+1357.10
0.14x+0.15x=4057.10
0.29x=4057.10
29x=405710
x=405710/29
x=$13,990 invested at 14%
$18,000-$13,990=$4010 invested at 15%
$13,990 invested at 14% and $4010 invested at 15%
check: 0.14*13,990=$1958.60
          0.15*4010=$601.50
$1958.60-$601.50=$1357.10
Let A1 be the account earning 14% and A2 be the account earning 15%
 
A1 = P1(ert) = P1(e(0.14*t))
 
A2 = P2(ert) = P2 (e(0.15*t))
 
We also know that A1 - A2 = $1357.1 and that P1 + P2 = $18,000.00
 
A1 = A2 + $1357.1  and  P1 = $18,000 - P2
 
P1(e(0.14*t)) = P2 (e(0.15*t)) + $1357.1
 
$1357.1 = P1(e(0.14*t)) - P2 (e(0.15*t))
 
$1357.1 = [$18,000 - P2](e(0.14t)) - P2 (e(0.15*t))
 
$1357.1 = ($18,000)(e(0.14t)) - P2[(e(0.14t)) + (e(0.15*t))]
 
I was able to find a solution for P2 by brute force calculation, and arrived at a figure of $8,368 so that P1 = $9,632.
 
All I can say, Enileda, is that, if this is truly an "Algebra 1" problem, then your instructor has a warped sense of humor.
 
 

Comments

Continuous compounding is fine, as long as you specify the time.  There is no solution to the problem of constant difference of interest earned over arbitrary time. 
I agree with Michael.  It troubled me greatly that no time was specified in the statement of the problem.  So, not knowing what else to do, I assumed one year.  The answer I gave above is true if and only if: (1) the interest is compounded continuously; and (2) if the elapsed time between opening the accounts and maturity is one year.  If either or both of these is not true, then my answers are hogwash and ought to be ignored.
 
William S.
Seeing the problem was from an Algebra I course, I assumed simple interest for one year.
A+B=18000
.14A=1357.10+.15B
 
A       +     B=18000
14A    -  15B=135710
 
15A   +  15B=270000
14A   -   15B=135710
29A             =405710
A=13990 and so B=4010
 
.14A=1948.60
.15B-=601.5
difference=1357.10