How much was invested in each account?

Question

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?

Answer 1

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

Answer 2

Let A₁ be the account earning 14% and A₂ be the account earning 15%

 

A₁ = P₁(e^rt) = P₁(e^(0.14*t))

 

A₂ = P₂(e^rt) = 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)) = P^₂ (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.

 

 

Answer 3

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