This is the complete question:

Michael invested $18000 in two different accounts. The account paying 14% annually earned $1357.10 more than the one earning 15% interest. How much was invested in each account?

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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 A_{1} be the account earning 14% and A_{2} be the account earning 15%

A_{1} = P_{1}(e^{rt}) = P_{1}(e^{(0.14*t)})

A_{2} = P_{2}(e^{rt}) = P_{2} (e^{(0.15*t)})

We also know that A_{1} - A_{2} = $1357.1 and that P_{1} + P_{2} = $18,000.00

A_{1} = A_{2} + $1357.1 and P_{1} = $18,000 - P_{2}

P_{1}(e^{(0.14*t)}) = P_{2} (e^{(0.15*t)}) + $1357.1

$1357.1 = P_{1}(e^{(0.14*t)}) - P_{2} (e^{(0.15*t)})

$1357.1 = [$18,000 - P_{2}](e^{(0.14t)}) - P_{2} (e^{(0.15*t)})

$1357.1 = ($18,000)(e^{(0.14t)}) - P_{2}[(e^{(0.14t)}) + (e^{(0.15*t)})]

I was able to find a solution for P_{2} by brute force calculation, and arrived at a figure of $8,368 so that P_{1} = $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.

Michael F. | Mathematics TutorMathematics Tutor

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

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