Question 67441
Let's say that x dollars were in the 8% account and the rest ($25000-x) was in the 9% account.
So Zoya Lon received two interest ammounts that added up to $2135.00 for the year.  You can set this up as follows using I = Prt: Where r1 = 8% or 0.08 and r2 = 9% or 0.09, P1 = x and P2 = $25000-x, and t = 1 year.
Note that two interest amounts I1 + I2 = $2135 

I1 = x(0.08)(1) = 0.08x
I2 = ($25000-x)(0.09)(1) = $2250-0.09x
Now add I1 and I2
I1+I2 = 0.08x + ($25000-x)(0.09) = $2135 Simplify this and solve for x.
0.08x+$2250-0.09x = $2135 
-0.01x+$2250 = $2135
-.01x = -$115 Divide both sides by -0.01
x = $11500 This is the amount invested at 8%
$25000-x = $25000-$11500 = $13500 This is the amount invested at 9%

Check:

0.08($11500) + 0.09($13500) = $920 + $1215 = $2135