Question 707181
let x = amount invested at 2%, y = amount invested at 3%


She invested $60,000 total, so


x+y = 60000


solve for y to get:  y = 60000 - x


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If she invests x dollars at 2% interest, then she will earn 0.02x dollars in interest alone after one year. Similarly, she will earn 0.03y dollars if she invests y dollars at 3%


In total, she will earn 0.02x + 0.03y dollars in interest alone from both accounts combined.


We're told that "In one year, she earned a total of $1600 in interest", so 


Total Interest Earned = 1600


0.02x + 0.03y = 1600


Now plug in  y = 60000 - x and solve for x


0.02x + 0.03y = 1600


0.02x + 0.03(60000 - x) = 1600


0.02x + 0.03(60000) - 0.03x = 1600


0.02x + 1800 - 0.03x = 1600


-0.01x + 1800 = 1600


-0.01x = 1600 - 1800


-0.01x = -200


x = -200/(-0.01)


x = 20000


So $20,000 was invested at 2% simple interest


y = 60000 - x


y = 60000 - 20000


y = 40000


and $40,000 was invested at 3% simple interest



Check: 


we can see that the two amounts $20,000 and $40,000 add to $60,000, so that checks out


$20,000 invested at 2% gives you 20000*0.02*1 = 400 dollars in interest alone


$40,000 invested at 3% gives you 40000*0.03*1 = 1200 dollars in interest alone


In total, you get 400+1200 = 1600 dollars in interest alone


So the answers are verified.