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?

1 Answer

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#