Question 319985
let $x be invested in first fund
let $y be invested in second fund

first fund  9% 
the second  3%, 
received 1047$. 
Last year interest equation 
0.09x+0.03y=1047
multiply by 100
9x+3y=104700 ...... equation 1
..
This year first fund  10% 
second  1% 
received  846$.  
This year interest equation
0.10x+0.01y=846
multiply by 100
10x+y=84600 ......... equation 2

multiply equation 2 by 3 and subtract1 from it
30x+3y-9x-3y=253800-104700
21x=149100
x= $ 7100 investment in first fund

.. 
10x+y=84600 substitute the value of x
10*7100+y=84600
y=84600-71000
y= $ 13600 investment in second fund.