invest $2,000, part earns 6% year and reh rest earned 11% year. Interest at end of year = $155. How much is invested at each rate?

To solve this problem, we can set up a system of equations based on the given information.

Let's assume that the amount invested at 6% is x dollars, and the amount invested at 11% is $2000 - x dollars (as the remaining amount is invested at 11%).

Now, let's calculate the interest earned at each rate:
Interest earned at 6% = x * 6% = (x/100) * 6 = 0.06x dollars
Interest earned at 11% = ($2000 - x) * 11% = (2000 - x)/100 * 11 = 22 - 0.11x dollars

According to the problem, the total interest earned at the end of the year is $155. So, we can set up the following equation:

0.06x + (22 - 0.11x) = 155

Now, let's solve this equation to find the value of x:

0.06x + 22 - 0.11x = 155
-0.05x + 22 = 155
-0.05x = 155 - 22
-0.05x = 133
x = 133 / -0.05
x = -2660

It seems like we have reached a negative value for x, which is not possible in this case. So, let's reassess the problem and see if there's any mistake.

It looks like we made an incorrect assumption about the amounts invested. Since investing a negative amount doesn't make sense, let's reset the equation:

Let x be the amount invested at 6% and y be the amount invested at 11%.

We know that x + y = $2000 (total amount invested)

The interest earned can be calculated as:
Interest at 6% = 0.06x
Interest at 11% = 0.11y

According to the problem, the total interest earned at the end of the year is $155. So, we can set up the following equation:

0.06x + 0.11y = 155

Now, let's solve this system of equations to find the values of x and y:

x + y = 2000 ---(Equation 1)
0.06x + 0.11y = 155 ---(Equation 2)

We can solve Equation 1 for x:
x = 2000 - y

Substituting x in Equation 2:
0.06(2000 - y) + 0.11y = 155
120 - 0.06y + 0.11y = 155
0.05y = 155 - 120
0.05y = 35
y = 35 / 0.05
y = 700

Now, we have found the value of y, which represents the amount invested at 11%, to be $700.

To find x, we can substitute the value of y in Equation 1:
x + 700 = 2000
x = 2000 - 700
x = 1300

Therefore, $1300 is invested at 6% and $700 is invested at 11%.