Mr. Oliver invested a total of $6400, part at 9% per year and the remainder at 8% per year. At the end of 2 years and 6 months, he had earced a total investment of $1380 in simple interest. How much ded he invest at each rate.
I have to come up with the equation.
Bex, I have to have 2 different numbers. I have to find out how much he invested at each individual rate.
I have .09x + .08(6400 - x)=1380.
This equation is not correct. Someone please help me!!
sorry.im not very good at algebra =[
i need 9 factors of 4x squared
To solve this problem, you need to set up two equations based on the given information.
Let's assume Mr. Oliver invested x dollars at 9% per year. Therefore, he invested (6400 - x) dollars at 8% per year.
Now, let's calculate the interest earned on each investment:
Interest earned at 9% = x * 0.09 * (2 + 6/12) = 0.09x * 2.5
Interest earned at 8% = (6400 - x) * 0.08 * (2 + 6/12) = 0.08(6400 - x) * 2.5
The total interest earned is given as $1380, so we can write the equation:
0.09x * 2.5 + 0.08(6400 - x) * 2.5 = 1380
Now, we can solve this equation to find the value of x.
0.225x + 0.2(6400 - x) = 1380
0.225x + 1280 - 0.2x = 1380
0.025x = 100
x = 4000
So, Mr. Oliver invested $4000 at 9% per year, and the remaining amount, $2400, was invested at 8% per year.
.09x+.08x(30)= $1380 + $6400 = ?
......i think/hope that this is right....
.09x+.08x(30)= $1380 + $6400 =
(Multiply .08 X 30 = 2.4)
.09x + 2.4x = 7780
3.3x = 7780
3.3x $7780
---- = -----
3.3 3.3
x = $2357.57