math
posted by jerson on .
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 > 2x2y=2000
[25] 0.08x+0.02y=310 >2x0.5y=7750
where and how did they get [25]?

They were trying to eliminate the decimals, but they did it in a very clumsy way
I would have multiplied the second by 100
8x + 2y = 31000
then the first by 2
2x  2y = 2000
trivial from here on ...
leaving a decimal after you multiplied by something sort of defeats the purpose, why didn't they just multiply the first by 50 and leave the first alone.
That would have done the trick also.