last year Jodie invested $10000, part at 6% anual intrest and the rest at 8% annual intrest. if she received $760 in intrest at the end of the year how much did she invest at each rate.
help is much appreciated plz help!!
x at 6% and 10,000-x at 8%
.06 x + .08 (10,000-x) = 760
.06 x + 800 -.08 x = 760
.02 x = 40
I think you can get x and 10,000 - x now
2000
To solve this problem, we can set up a system of equations. Let's say Jodie invested x dollars at 6% interest and (10000 - x) dollars at 8% interest.
The interest earned from the investment at 6% can be calculated as: 0.06x (since 6% can be written as 0.06).
The interest earned from the investment at 8% can be calculated as: 0.08(10000 - x) (since 8% can be written as 0.08).
According to the problem, the total interest earned is $760. So, we can set up the equation:
0.06x + 0.08(10000 - x) = 760
Now, let's solve the equation step-by-step:
0.06x + 0.08(10000 - x) = 760
0.06x + 800 - 0.08x = 760
-0.02x = 760 - 800
-0.02x = -40
Next, we need to isolate x by dividing both sides of the equation by -0.02:
x = (-40) / (-0.02)
x = 2000
Therefore, Jodie invested $2000 at 6% interest. To find the amount invested at 8%, we subtract this from the total investment:
10000 - x = 10000 - 2000 = 8000
So, Jodie invested $2000 at 6% interest and $8000 at 8% interest.