Part of 26000 invested at 4% interest and the rest at 2% interest yields 680.00
Invested $X @ 4%.
Invested $(26,000 - X) @ 2%.
0.04X + 0.02(26000 - X) = 680,
0.04X + 520 - 0.02X = 680,
0.02X = 680 - 520 = 160,
X = $8000 @ 4%.
26000 - 8000 = $18000 @ 2%.
To solve this problem, we can use a system of equations.
Let's assume that the amount invested at 4% interest is x, and the amount invested at 2% interest is y.
According to the problem, the total amount invested is $26,000. Therefore, we can write the equation:
x + y = 26,000 -- Equation 1
The problem also states that the total interest earned from the investments is $680. We can calculate the interest earned from each amount using the formula: Interest = Principal × Rate × Time.
For the amount invested at 4% interest, the equation becomes:
0.04x
For the amount invested at 2% interest, the equation becomes:
0.02y
Now, we can set up the second equation using the above information:
0.04x + 0.02y = 680 -- Equation 2
To solve this system of equations, we can use substitution or elimination method. Let's use substitution:
From Equation 1, we can express x in terms of y: x = 26,000 - y
Substituting the value of x in Equation 2, we get:
0.04(26,000 - y) + 0.02y = 680
Now, we can solve for y:
1,040 - 0.04y + 0.02y = 680
1,040 - 0.02y = 680
-0.02y = 680 - 1,040
-0.02y = -360
y = -360 / -0.02
y = 18,000
Now, substitute the value of y back into Equation 1 to solve for x:
x + 18,000 = 26,000
x = 26,000 - 18,000
x = 8,000
Therefore, $8,000 was invested at the 4% interest rate, and $18,000 was invested at the 2% interest rate to yield $680 in total interest.