SOLUTION: You invest $6000 in two accounts paying 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same interest. How much was invested at the same rat

Question 96926This question is from textbook Introductory Algebra (Blitzer)
: You invest $6000 in two accounts paying 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same interest. How much was invested at the same rate? This question is from textbook Introductory Algebra (Blitzer)

Answer by mathslover(157) (Show Source): You can put this solution on YOUR website!
let the amount invested in the account paying 6% be A.
Therefore amount invested in the account paying 9% is 6000 -A
Interest earned from the 6% account in 1 year = 6/100 * A
Interest earned from the 9% account in 1 year = 9/100 * (6000-A)
Since both the interests are same
=> 6/100 * A = 9/100 * (6000-A)
Multiplying both sides by 100
6A = 9( 6000-A)
6A = 54000 - 9A
Transposing the similar terms at one side
15 A = 54000
therefore A= 3600
Amount invested at 6% = $3600
therefore, amount invested at 9% = $6000 - $3600 = $2400

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