260 algebra De mystif ieD
HINt If we let x represent the amount earned at interest rate A, then the rest is
invested at interest rate B. “The rest” is the total dollars less the amount invested
at interest rate A. If x dollars is invested at interest rate A, then the amount of inter-
est earned in a year is Ax. If “Total – x” dollars is invested at interest rate B, then
B(Total – x) is the amount of interest earned in a year at interest rate B. We use the
following equation (model) to solve these problems.
Ax (+ BxTotal – ) = InterestearnedEXAMPLE
Dora had $10,000 to invest. She deposited her money into two accounts—
one paying 6% interest and the other 721 % interest. If at the end of the year
the total interest earned was $682.50, how much was originally deposited
in each account?
We could either let x represent the amount deposited at 6% or at 712 %. Here,
we will let x represent the amount deposited into the 6% account. Because
the two amounts must sum to 10,000, 10,000 – x is the amount deposited at
712 %. The amount of interest earned at 6% is 0.06x, and the amount of inter-
est earned at 712 % is 0.075(10,000 – x). The total amount of interest is
$682.50, so we want to solve 0.06x + 0.075(10,000 – x) = 682.50. The solution
to this equation gives us the amount to invest at 6%, and putting this solu-
tion into 10,000 – x gives us the amount to invest at 712 %.0060 075 10 000 682 50
006 750 0 075 68..(, ).
..xx
xx+−=
+− = 2250
0 015 750 682 50
750 750 00
0 015 675.
..
.
..−+=
−−
−=−xx 00
67 50
0 015
4500xx=−
−
=.
.Dora deposited $4500 in the 6% account and 10,000 – x = 10,000 – 4500 =
$5500 in the 721 % account.EXAMPLE
Dora had $10,000 to invest. She deposited her money into two accounts—