Final answer:
To solve for the amounts invested, we use two key equations: one to represent the total investment of $6,800 and another for the total interest of $436. By solving these equations, we find that Russell must have invested $1,400 at 8% interest and $5,400 at 6%. However, none of the provided answer choices match these calculated amounts, suggesting a potential error in the problem's details.
Step-by-step explanation:
Russell has invested $6,800 across two different accounts, one yielding 8% interest, and the other 6%. After one year, he earned $436 in total interest from both accounts. To ascertain the amount invested in each, we can set up the following system of linear equations:
- Let x be the amount invested at 8%.
- Let y be the amount invested at 6%.
The total amount invested is the sum of x and y, which equals $6,800.
x + y = 6800 (Equation 1)
The total interest earned from both accounts is $436, which is the sum of the interests from each investment:
0.08x + 0.06y = 436 (Equation 2)
To solve this system, we can multiply Equation 2 by 100 to clear the decimals and then use substitution or elimination methods. By multiplying Equation 2, we get:
8x + 6y = 43600
Now, if we multiply Equation 1 by 6 to align the terms, we have:
6x + 6y = 40800 (Equation 3)
Subtracting Equation 3 from the modified Equation 2 yields:
2x = 2800
Divide both sides by 2 to find x:
x = 1400
Now, we can calculate y by substituting x back into Equation 1:
1400 + y = 6800
y = 6800 - 1400
y = 5400
Therefore, Russell invested $1,400 at 8% and $5,400 at 6%. Unfortunately, none of the choices provided (a, b, c, d) match the correct amounts we've calculated. There might be an error in the statement of the problem or in the provided choices.