Helene invested a total of $1,000 in two simple-interest bank accounts. One account paid 6% annualinterest; the other paid 7% annual interest. The total amount of interest she earned after one year was$66. Enter and solve a system of equations to find the amount invested in each account. Enter the interestrates in order as given in the problem. (Hint: Change the interest rates into decimals first.)x+y = 1,0000.06x+ 0.07y = 66Helene invested $at 6% and $at 7%.​

Answer:Helene invested $ 400 at 6% and $ 600 at 7%.​Step-by-step explanation:Letx -----> the amount invested at 6%y -----> the amount invested at 7%we know thatThe interest earned with the amount invested at 6% plus the interest earned by the amount invested at 7% must be equal to $66Remember that[tex]6\%=6/100=0.06\\7\%=7/100=0.07[/tex]so[tex]x+y=1,000[/tex] ----> equation A[tex]0.06x+0.07y=66[/tex] ----> equation BSolve the system of equations by graphingThe solution of the system is the intersection point both graphsusing a graphing toolThe solution is the point (400,600)see the attached figurethereforeHelene invested $ 400 at 6% and $ 600 at 7%.​