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.
Nov 19

William S.
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