A woman invested some money at 8% and some at 9% annual simple interest. The interest for 1 year on the combined investment of $10,000 was $860. How much was invested at each rate?
If $x is at 8%, then
.08x + .09(10000-x) = 860
$2000 at 8% & $8000 at 9% interest was invested.
Racquel, check your calculations.
2000(.08) + 8000(.09) = 880 , not 860 as needed.
To solve this problem, we need to set up two equations based on the given information.
Let's say the amount invested at 8% is x (in dollars), and the amount invested at 9% is y (in dollars). We know that the combined investment is $10,000, so we have the equation:
x + y = 10,000 ----(Equation 1)
We also know that the interest earned from the investment for one year is $860. The interest earned from investing x dollars at 8% is 0.08x, and the interest earned from investing y dollars at 9% is 0.09y. Therefore, we have the equation:
0.08x + 0.09y = 860 ----(Equation 2)
Now, we can solve these two equations simultaneously to find the values of x and y.
Method 1: Substitution Method
1. Solve Equation 1 for x: x = 10,000 - y
2. Substitute the value of x in Equation 2: 0.08(10,000 - y) + 0.09y = 860
3. Simplify the equation: 800 - 0.08y + 0.09y = 860
4. Combine like terms: 0.01y = 60
5. Solve for y: y = 60 / 0.01 = 6,000
6. Substitute the value of y back into Equation 1 to find x: x = 10,000 - 6,000 = 4,000
Therefore, $4,000 was invested at 8% and $6,000 was invested at 9%.
Method 2: Elimination Method
1. Multiply Equation 1 by 0.08 to eliminate x: 0.08x + 0.08y = 800
2. Subtract the new equation from Equation 2 to eliminate x: (0.08x + 0.09y) - (0.08x + 0.08y) = 860 - 800
3. Simplify the equation: 0.01y = 60
4. Solve for y: y = 60 / 0.01 = 6,000
5. Substitute the value of y back into Equation 1 to find x: x = 10,000 - 6,000 = 4,000
Therefore, $4,000 was invested at 8% and $6,000 was invested at 9%.