Question

Peter invested some money at 6% annual interest, and Martha invested some at 12%. If their combined investment was $6,000 and their combined interest was $450, how much money did Martha invest?

Answer 1

Peter invested `$.4500`
Martha invested `$.1500`

Explanation:

Peter invested `$.x`
Martha invested `$.y`

Interest from `$.x= x xx 6/100=(6x)/100`

Interest from `$.y= y xx 12/100=(12y)/100`

Then -

`(6x)/100+(12y)/100=450`

To do away with fraction, let us multiply both sides by 100

`6x+12y=45000`----------(1)

`x+y=6000`-----------------(2)

Let us solve the 2nd equation for `x`

`x=6000-y`

Plug in the value of `x=6000-y` in equation (1)

`6(6000-y)+12y=45000`
`36000-6y+12y=45000`
`6y=45000-36000=9000`
`y=9000/6=1500`

Substitute `y=1500` in equation (2) and simplify

`x+1500=6000`
`x=6000-1500=4500`

Peter invested `$.4500`
Martha invested `$.1500`