A person invested $7900 for one year, part at 5%, part at 11%, and the remainder at 15%. The total annual income from these investments was $965. The amount of money invested at 15% was $500 more than the amounts invested at 5% and 11% combined. Find the amount invested at each rate.

To solve this problem, let's break it down into smaller steps.

Step 1: Let's represent the amount invested at 5% as 'x'.
Step 2: Since the amount invested at 11% is the same as the amount invested at 5% (according to the problem), the amount invested at 11% is also 'x'.
Step 3: According to the problem, the amount invested at 15% is $500 more than the amounts invested at 5% and 11% combined. So, the amount invested at 15% is 'x + x + 500'.
Step 4: Now let's calculate the total investment. The total investment is the sum of the amounts invested at each rate: 'x + x + x + 500 = 7900'. Simplifying this equation, we get '3x + 500 = 7900'.
Step 5: To isolate 'x', subtract 500 from both sides of the equation: '3x = 7400'.
Step 6: Finally, divide both sides of the equation by 3 to solve for 'x': 'x = 7400/3'.

Therefore, the amount invested at 5% is $2466.67 (approx.), and the amount invested at 11% is also $2466.67 (approx.). The amount invested at 15% is $5433.33 (approx.).

Step 1: Let's begin by defining our variables:

Let x be the amount invested at 5%.
Let y be the amount invested at 11%.
Let z be the amount invested at 15%.

Step 2: We know that the total amount invested is $7900, so we have:
x + y + z = 7900 (equation 1)

Step 3: We also know that the total annual income from these investments was $965, so we have:
0.05x + 0.11y + 0.15z = 965 (equation 2)

Step 4: According to the problem, the amount invested at 15% was $500 more than the amounts invested at 5% and 11% combined. This can be expressed as:
z = x + y + 500 (equation 3)

Step 5: Now we have a system of three equations with three variables. We can solve this system using substitution or elimination. Let's use substitution to solve for the variables.

Step 6: From equation 3, we can substitute the value of z in equations 1 and 2:
x + y + (x + y + 500) = 7900
2x + 2y + 500 = 7900
2x + 2y = 7400 (equation 4)

0.05x + 0.11y + 0.15(x + y + 500) = 965
0.05x + 0.11y + 0.15x + 0.15y + 75 = 965
0.2x + 0.26y = 890 (equation 5)

Step 7: Multiply equation 4 by -0.13 and add it to equation 5 to eliminate the x variable:
-0.26x - 0.26y = -962
0.2x + 0.26y = 890
-0.06x = -72
x = 1200

Step 8: Substitute the value of x into equation 4 to solve for y:
2(1200) + 2y = 7400
2400 + 2y = 7400
2y = 5000
y = 2500

Step 9: Substitute the values of x and y into equation 1 to solve for z:
1200 + 2500 + z = 7900
3700 + z = 7900
z = 4200

Step 10: Therefore, the amount invested at each rate is $1200 at 5%, $2500 at 11%, and $4200 at 15%.

x+y+z = 7900

.05x + .11y + .15z = 965
z = x+y + 500