SOLUTION: I need help setting up the following problem: You invest $6000 in two accounts payinhg 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same

Algebra ->  Finance -> SOLUTION: I need help setting up the following problem: You invest $6000 in two accounts payinhg 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same       Log On


   

Question 102484This question is from textbook Introductory Algebra for College Students
: I need help setting up the following problem:
You invest $6000 in two accounts payinhg 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same interest. How much was invested at each rate? This question is from textbook Introductory Algebra for College Students

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I need help setting up the following problem:
You invest $6000 in two accounts payinhg 6% and 9% annual interest, respectively. At the end of the year, the accounts earn the same interest. How much was invested at each rate?
Let's say you invest X at 6% and Y at 9%.
You know that
1.X%2BY=6000
You also know that the interest for each is the same
I%5Bx%5D=X%2A6%2F100=INT
I%5By%5D=Y%2A9%2F100=INT
Set them equal to each other.
X%2A6%2F100=Y%2A9%2F100
2.X%2A6=Y%2A9
Now you have two equations with two unknowns (X and Y).
Use Equation (2) and solve for X.
X=9%2AY%2F6=3%2AY%2F2
Substitute into Equation (1)
1.X%2BY=6000
3%2AY%2F2%2BY=6000Substitute
3%2AY%2B2%2AY=12000Multiply both sides by 2.
5%2AY=12000Simplify.
cross%285%29%2AY%2Fcross%285%29=12000%2F5Multiplicative inverse of 5.
Y=2400
Use equation (1)
X%2BY=6000
X=3600
$3600 invested at 6%.
$2400 invested at 9%.
$216 made by each account.