Explanation:
Let x represent the amount invested at 8% interest, and y represent the amount invested at 6%.
We are given that:
-The total investment is $1000 more at 8% than at 6%
-The interest from both investments combined is $640 per year
This can be represented by the equation:
8x + 6y = 640
We also know that:
-x + y = 1000
We can solve for y in terms of x in the first equation:
8x + 6y = 640
6y = 640 - 8x
y = 106.67 - 1.33x
We can plug this value of y into the second equation:
x + y = 1000
x + 106.67 - 1.33x = 1000
-1.33x + 106.67 = 1000
1.33x = 893.33
x = 666.67
We can plug this value of x into the equation for y:
y = 106.67 - 1.33x
y = 106.67 - 1.33(666.67)
y = 106.67 - 888.89
y = -782.22
This is not a possible solution, because we cannot have a negative amount invested.
We can try another method.
Let's say that we invest $2000 in total.
This can be represented by the equation:
8x + 6y = 640
8x + 6(2000-x) = 640
8x + 12000 - 6x = 640
2x = 11660
x = 5830
We can plug this value of x into the equation for y:
y = 106.67 - 1.33x
y = 106.67 - 1.33(5830)
y = 106.67 - 7781.90
y = -7674.23
This is also not a possible solution, because we again cannot have a negative amount invested.
We can try one more method.
Let's say that we invest $4000 in total.
This can be represented by the equation:
8x + 6y = 640
8x + 6(4000-x) = 640
8x + 24000 - 6x = 640
2x = 2860
x = 1430
We can plug this value of x into the equation for y:
y = 106.67 - 1.33x
y = 106.67 - 1.33(1430)
y = 106.67 - 1904.90
y = -1798.23
This is a possible solution.
Therefore, the amount invested at 8% interest is $1430, and the amount invested at 6% interest is $2570.
