Larry Mitchell invested part of his $31,000 advance at 6% annual simple interest and the rest at 7% annual simple interest. If the total yearly interest from both accounts was $2,090, find the amount invested at each rate.
Let x be the amount invested at 6%. Then 31000 - x is invested at 7%.
We have the following equation:
0.06x + (31000 - x)0.07 = 2090
Simplify:
0.06x + 2170 - 0.07x = 2090
Combine like Terms
-0.01x + 2170 = 2090
Subtract 2170 from each side
-0.01x = -80
Divide each side by -0.01
x = 8000 at 6%
Which means at 7%, we have:
31000 - 8000 = 23,000
Let x be the amount invested at 6%. Then 31000 - x is invested at 7%.
We have the following equation:
0.06x + (31000 - x)0.07 = 2090
Simplify:
0.06x + 2170 - 0.07x = 2090
Combine like Terms
-0.01x + 2170 = 2090
Subtract 2170 from each side
-0.01x = -80
Divide each side by -0.01
x = 8000 at 6%
Which means at 7%, we have:
31000 - 8000 = 23,000