Homework help : interest problem!!?

Question

a man invested $6000, part of it at 5% simple interest and the rest at 7% simple interest. if his annual interest income is $372, how much did he invest at each rate?

hopefully someone can tutor me on this one! i done research online and stuff but i still don't get it!!

Answer 1

lucky u have mE!, dont know how i know this  :O

Its probaly wrong though :P

But i think its SI=PRT/100

SI= Simple Interest

P=Price

R= Rate

T=Time

Answer 2

Hmmm actauly i think u should do wat that guy said lol

Answer 3

.05x+.07y=372

x+y=6000

solve for x in one of equations first, then pug it into the other to solve for y.

x=6000-y

.05(6000-y)+.07y=372

300-.05y+.07y=372

300+.02y=372

.02y=72

y=3600

x+y=6000

x+3600=6000

x=2400

2400 at 5%, 3600 at 7%

Answer 4

5%x + 7%y = 372

x + y = 6000 ---> y = 6000 - x

====> 5/100*x + 7/100(6000 - x) = 372

====> 5/100*x + 420 - 7/100*x = 372

===> -2*x/100 =  -48

===> x = $2400

===> y = $3600

Answer 5

let x represent the amount invested at 5% and y be the amount invested at 7%.

x + y = 6000

0.05x + 0.07y = 372

0.05x + 0.05y = 300

-(0.05x + 0.07y = 372)

-0.02y = - 72

y = 3600

sub into x + y = 6000

x + 3600 = 6000

x = 6000 - 3600

x = 2400

$2400 was invested at 5% and $3600 was invested at 7%.  hope this helps! this is 'elimination' btw.