Doris invested some money at 7% and some money at 8%.She invested $6000 more than she did at 7%. Her total yearly interest from the two investments was $780. How much did Doris invest at each rate?
If she invested $x at 7%, then if I understand what you wrote, she invested $(x+6000) at 8%, right?
Now add up the interest
.07x + .08(x+6000) = 780
To solve this problem, let's assign variables to the unknowns.
Let x be the amount that Doris invested at 7%.
Since she invested $6000 more at 8% than she did at 7%, she invested (x + $6000) at 8%.
Now, we can calculate the interest from each investment.
The interest from the investment at 7% is calculated as:
0.07x (7% expressed as a decimal)
The interest from the investment at 8% is calculated as:
0.08(x + $6000) (8% expressed as a decimal)
According to the problem, the total interest from the two investments is $780. Therefore, we can set up the equation:
0.07x + 0.08(x + $6000) = $780
Let's solve this equation:
0.07x + 0.08x + 0.08($6000) = $780
0.07x + 0.08x + $480 = $780
0.15x + $480 = $780
0.15x = $780 - $480
0.15x = $300
To isolate x, divide both sides of the equation by 0.15:
x = $300 / 0.15
x = $2000
So, Doris invested $2000 at 7% and $2000 + $6000 = $8000 at 8%. Therefore, she invested $2000 at 7% and $8000 at 8%.