A total of $11,000 was invested. Part of the $11,000 was invested at 4% and the rest invested at 7%. If the investments earn $680 per year, how much was invested at each rate?
(x)(.04)+(11000-x)(.07)=680
where x=the amount of money invested at 4%
and 11000-x = the rest of the 11000 invested at 7%
4800
A total of $11,000 was invested .Part of the $11,000 was invested at 4 percent and the rest was invested 7percent if the investment earn $680 per year how much was invested at each rate?
To solve this problem, we can use a system of equations.
Let's say the amount invested at 4% is x dollars. Therefore, the amount invested at 7% would be (11000 - x) dollars.
Now we can set up two equations based on the given information:
0.04x + 0.07(11000 - x) = 680 (equation 1 - represents the total interest earned)
x + (11000 - x) = 11000 (equation 2 - represents the total amount invested)
Now, let's solve this system of equations:
From equation 2, we get x + 11000 - x = 11000.
Combining like terms, we have 11000 = 11000, which is always true. This equation doesn't give us any specific value for x but confirms that it is a valid solution.
Now, let's substitute this result back into equation 1:
0.04x + 0.07(11000 - x) = 680
0.04x + 770 - 0.07x = 680
Combining like terms, we have -0.03x + 770 = 680
Subtracting 770 from both sides, we get -0.03x = -90
Dividing both sides by -0.03, we find x = 3000.
Therefore, $3000 was invested at 4% and $8000 (11000 - 3000) was invested at 7%.
In summary, $3000 was invested at 4% and $8000 was invested at 7%.