A total of $15,000 is invested into two accounts. One of the accounts earns 12% per year and the other earns 10% per year. If the total interest earned in the first year is $1600, how much is invested in each account?
1.) Define the variable(s) and what they stand for.
2.) Set up an equation
3.) State the solutions
amount invested at 12% --- x
amount invested at 10% --- 15000-x
.12x + .10(15000-x = 1600
12x + 10(15000-x) = 160000
12x + 150000 - 10x = 160000
2x = 10000
x = 5000
so $5000 invested at 12% and $10000 invested at 10%
check:
.12(5000) + .1(10000) = 1600
YUP!
12
1.) Let's define the variables:
Let x be the amount invested in the account earning 12% per year.
Let y be the amount invested in the account earning 10% per year.
2.) Set up an equation:
The interest earned from the account earning 12% per year is given by the formula: 0.12x.
The interest earned from the account earning 10% per year is given by the formula: 0.10y.
The total interest earned in the first year is $1600.
So, the equation will be: 0.12x + 0.10y = $1600.
3.) State the solutions:
We have one equation with two unknowns, so we need more information to solve it.
1.) Let's define the variables:
Let x be the amount invested in the account earning 12% per year.
Let y be the amount invested in the account earning 10% per year.
2.) Setting up the equation:
For the amount invested in the first account, the interest earned can be calculated as: 0.12x
For the amount invested in the second account, the interest earned can be calculated as: 0.10y
According to the given information, the total interest earned in the first year is $1600. This can be represented as: 0.12x + 0.10y = $1600
Additionally, it is known that the total amount invested is $15,000. This can be represented as: x + y = $15,000
3.) State the solutions:
Now we have a system of two equations:
0.12x + 0.10y = $1600
x + y = $15,000
Solving this system of equations will give us the solution for x and y, which represents the amounts invested in each account.