tracy invested $6000 for one year, part at 10% annual and the balance at 13% annual interest. her total interest for the year was $712.5. how much money did she invested at each rate?
I tried using a table for this I just don't know how to set the problem up.
Let $x is what she invested at 10%
then $(6000-x) is what she invested at 13%.
We know that
0.10x + 0.13(6000-x) = 712.5
Solve for x to get the answers.
To solve this problem, we can set up a system of two equations with two variables.
Let's first consider the amount of money Tracy invested at 10% annual interest. Let's call this amount "x".
Since Tracy invested a total of $6000, the amount invested at 13% interest can be represented as $6000 - x.
Now, we can calculate the interest earned from each investment. The interest earned from the amount invested at 10% interest can be calculated as 0.10x (10% = 0.10 as a decimal).
Similarly, the interest earned from the amount invested at 13% interest can be calculated as 0.13(6000 - x).
According to the given information, Tracy earned a total interest of $712.5. So, we can set up the following equation:
0.10x + 0.13(6000 - x) = 712.5
Now, we can solve this equation to find the value of "x" (the amount invested at 10% interest) and then calculate the amount invested at 13% interest.
Let's solve the equation step by step:
0.10x + 0.13(6000 - x) = 712.5
0.10x + 780 - 0.13x = 712.5 (distributing 0.13)
Combine like terms:
0.10x - 0.13x + 780 = 712.5
-0.03x + 780 = 712.5 (simplifying)
Subtract 780 from both sides:
-0.03x = 712.5 - 780
-0.03x = -67.5
Now, divide both sides by -0.03:
x = (-67.5) / (-0.03)
x ≈ 2250
So, Tracy invested approximately $2250 at 10% interest.
To find the amount invested at 13% interest, subtract x from the total investment:
6000 - x = 6000 - 2250 = $3750
Thus, Tracy invested $2250 at 10% interest and $3750 at 13% interest to earn a total interest of $712.5.