Angus (was he a bull? J) invested $18,000, part at 3%
and part at 5%.  If the total interest at the end of 
the year is $660, how much did he invest at each 
rate?

Let $x represent the first part, that is, the amount 
invested at 3%

Now use the formula for "the rest"

æThe rest of   ö    æ The  ö   æ the ö
ça total amount÷  = çtotal ÷ - çfirst÷
çbesides the   ÷    èamountø   è partø
èfirst part    ø	

So the rest of the $18000 besides the $x
is = ($18000 - $x)

This amount (the rest besides the $x) was invested
at 5%. 

So we have x dollars invested at 3%
and (18000 - x) dollars invested at 5%

The interest on the x dollars is .03x dollars.

The interest on the (18000 - x) dollars is
.05(18000 - x)

We read:
>>...the total interest at the end of the year 
is $660...<<

|Interest |     |Interest|
|   on    |     |   on   |
|  the    |  +  |  the   |  =  $660 
|x dollars|     |18000-x |
                |dollars |
       
     .03x    +  .05(18000-x) = 660

Can you solve that? If not post again asking how.

You get x = $12000, which is the amount invested
at 3%.

To get the rest, which was invested at 5%, subtract
that $12000 from the total, $18000, and get $6000, 
the amount invested at 5%.

Edwin