a trustee invested a total of $100,000, more than half of which was invested at 6 percent simple interest and the rest of which was invested at 8 percent simple annual interest. what amount was invested at 8 percent?
1) the total annual income on the investment was $6800
2) if twice as much had been invested at 8 percent and the rest at 6 percent, the total annual income on the investment would have been increased by $800

Say x is amount more than half.
Therefore 50000+x was invested in 6% => ((50000+x)*6)/100 .....(1)
And 50000-x was invested in 8% => ((50000-x)*8)/100 ....(2)

Statement 1
((50000+x)*6)/100 + ((50000-x)*8)/100 = 6800
So X can be calculated. So we can easily say that Statement 1 is sufficient. As correctly pointed out by you.

what will be the equation for statement 2?

Statement 2

If twice as much had been invested.

The original amount invested at 8% is ((50000-x)*8)/100 . So twice as much would be
((2(50000-x))*8)/100
And the rest to be invested at 6% would be ((100000 - (2(50000-x)))*6)/100.

((2(50000-x))*8)/100 + ((100000 - (2(50000-x)))*6)/100 = (((50000+x)*6)/100 + ((50000-x)*8)/100) + 800

So its again a linear eq which can be solved fro x

From Statement 1:

(6x)/100 + (1,00,000-x)8/100 = 6800

Solving this linear equation for x, we have x=60,000
So, the two sums are 60,000 and 40,000

Second equation:
40000*2=80,000 at 8%
80000*8/100 + 20000*6/100 = 7600 which is $ 800 more than the first income.