Larry Mitchell invested part of his $40,000 advance at 6% annual simple interest and the rest at 4% annual simple interest. If his total yearly interest from both accounts was $2,340, find the amount invested at each rate.
+129840 Let the amount invested at 6% = A......so the amount invested at 4% must be 40,000-A
So we have
A * .06 + (40,000 - A) * .04 = 2340 simplify
.06A - .04A + 1600 = 2340
.02A + 1600 = 2340 subtract 1600 from both sides
.02A = 740 divide both sides by ,02
A = 37,000 and this is the amount at 6%
And the amount at 4% = 40000 - A = 40000 - 37000 = 3000
