Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $1000 more invested at 8% than at 2%. Find the amount invested at each rate if the annual interest from both investments is $310. Let x amount invested at 8% and y = amount invested at 2%. Then the system that models the problem is [x=y+1000

Question

Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $1000 more invested at 8% than at 2%. Find the amount invested at each rate if the annual interest from both investments is $310. Let x amount invested at 8% and y = amount invested at 2%. Then the system that models the problem is [x=y+1000
                   0.08x+0.02y=310 Solve the system using the method of addition?=.

ok this is how my book says solve it
[2] x - y=100 -----> 2x-2y=2000
[-25] 0.08x+0.02y=310 -->-2x-0.5y=-7750

where and how did they get [-25]?