You invested 29,000 in two accounts paying 6% and 8% annual interest, respectively. if the total interest earned for the year is $2060, how much was invested at each rate

You can solve this using a system of equations:
0.06x + 0.08y = 2060 (interest)
x + y = 29,000 (money invested)

Multiply each term in the second equation by -0.06 to cancel out the 0.06x from the first equation.

-0.06x - 0.06y = -1740

Now add this equation to the first equation and solve for y.

 0.06x + 0.08y = 2060
-0.06x - 0.06y = -1740

0.02y = 320
y = 16,000

Since you know y = 16,000 you can substitute it into either of the original equations to solve for x. I picked the second equation because it's simpler, but you could also use the first.

x + 16,000 = 29,000
x = 13,000

So you invested 16,000 at 8% and 13,000 at 6%.

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