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Question

Larry MItchell invested part of his 14000$ advance at 5% annual simple interest and the rest at 2% annual interest. If his total yearly interest from both accounts was $610, find the amount invested at each rate

Answer 1

a + b = 14,000

.05a + .02b = 610

 

Solve the first for a or b. Let's do a.

 

a = 14,000 - b

 

Substitute!

 

.05(14,000 - b) + .02b = 610

 

Solve for b, then plug it back into the first equation to find a.