Sam invested $2150, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $22 more than twice the income from the 6% investment. How much did he invest at each rate?

To solve this problem, let's break it down step by step.

Step 1: Assign variables.
Let's assume that Sam invested x dollars at 6% and the remaining (2150 - x) dollars at 8%.

Step 2: Write equations based on the given information.
- The yearly income on the 8% investment was $22 more than twice the income from the 6% investment. We can write this as:
0.08 * (2150 - x) = 2 * (0.06 * x) + 22

Step 3: Solve the equation.
Let's solve the equation to find the value of x.

0.08 * (2150 - x) = 0.12 * x + 22
0.08 * 2150 - 0.08 * x = 0.12 * x + 22
172 - 0.08x = 0.12x + 22
-0.08x - 0.12x = 22 - 172
-0.2x = -150
x = -150 / -0.2
x = 750

Step 4: Calculate the remaining investment.
Sam invested $750 at 6%, so the remaining investment at 8% is:
2150 - 750 = $1400

Therefore, Sam invested $750 at 6% and $1400 at 8%.

Let's assume that Sam invested x dollars at 6% interest rate.

Therefore, the amount invested at 8% interest rate would be (2150 - x) dollars.

The income from the 6% investment would be (x * 6/100) dollars.
And the income from the 8% investment would be ((2150 - x) * 8/100) dollars.

According to the problem, the income from the 8% investment is $22 more than twice the income from the 6% investment. So, we can set up the following equation:

((2150 - x) * 8/100) = 2 * (x * 6/100) + 22

Now we can solve this equation to find the value of x.

if $x at 6%, the rest (2150-x) is at 8%. So, now we have

.08(2150-x) = 22+2(.06x)

now just crank it out.