PLEASE HELP. SIMPLE INTEREST PROBLEM.
A woman invested $8000 into mutual funds in one year part of your investment earned 7% simple interest and the other part earned 6% simple interest. At the end of the year the woman had earned $512. how much did she invest at each rate?

Question

PLEASE HELP. SIMPLE INTEREST PROBLEM.
A woman invested $8000 into mutual funds in one year part of your investment earned 7% simple interest and the other part earned 6% simple interest. At the end of the year the woman had earned $512. how much did she invest at each rate?

anonymous · Answer

i got .07x+.06y=512

anonymous · Answer

and x+y=8000

anonymous · Answer

She invested x amount with 6% and y amount with 7%
We know she had 8000 cash, so x+y=8000
The interest with 6% is 0.06 that is 0.06x , same for y 0.07y
They totaled a profit of 512 thus
0.06x+0.07y=512

anonymous · Answer

i just am not sure what to do from there

anonymous · Answer

That is good :)

anonymous · Answer

now i use substitution or elimination? idk how

anonymous · Answer

Well just solve them :) It looks a bit nasty

anonymous · Answer

I DONT KNOW HOW TO SOLVE THEM

anonymous · Answer

x+y=8000
.07x+.06y=512   Multiply by 100 to simplify
7x+6y=51200

First equation multiply by 6
6x+6y=48000
Take away second from first

7x+6y=51200
minus 
6x+6y=48000

x=3200

Do not yell at someone ever!! Especially when they are helping you..... really rude

anonymous · Answer

lmao i put it in caps by accident I'm sorry! I'm not yelling at you!! i appreciate your help, honestly.

anonymous · Answer

OK, never mind. 
This was the method of elimination. 
I made 6y for both equations so that I could take away one from the other

anonymous · Answer

thankyou so much sir i understand now