Answer:
$2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.
Step-by-step explanation:
Let a, b, and c be the fractions of the $12,000 that is invested in the savings, bond, & money fund respectively.
Let S1, S2, S3 be the total monies from the savings, bond, & money fund after the year with interest respectively.
We know that a+b+c=1
and
S1+S2+S3 = 12000 + 860
S1+S2+S3 = 12860
From, the question, we know that b = 2a, and c = 1 - 3a based on the condition of 2 times more in the bond account that the savings account.
Thus;
S1 = a(12000)(1.04) = 12480a
S2 = 2a(12000)(1.06) = 25,440a
S3 = (1 - 3a)(12000)(1.09) = 13080 - 39240a
We already know that;
S1+S2+S3 = 12860
Thus, Adding the 3 equations gives;
12480a + 25,440a + 13080 - 39240a = 12860
Thus, simplifying gives;
13080 - 12860 = 39240a - 12480a - 25,440a
220 = 1320a
a = 220/1320
a = 0.1667
So,b = 2 x 0.1667 = 0.3334
c = 1 - (3 x 0.1667)
c = 1 - 0.5
c = 0.5
Thus,
S1 = 12480a = 12480 x 0.1667 ≈ $2080
S2 = 25,440a = 25440 x 0.1667 ≈ $4241
S3 = 13080 - 39240(0.1667) ≈ 6539
So, $2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.
