Given:
a.) A person had 14,000 infested in two accounts.
b.) One paying 9% simple interest.
c.) One paying 10% simple interest.
Let,
x = the amount invested at 9% simple interest
y = the amount invested at 10% simple interest
1.) We know the total amount of money invested is $14,000. We get,
x + y = 14,000
2.) We know that the total interest for the year for the two accounts is $1432. We get,
0.09*x + 0.1*y = 1,339
Let's equate the two equations,
x = 14,000 - y (Substitute for x)
0.09*(14,000 - y) + 0.1*y = 1,339
1,260 - 0.09y + 0.1y = 1,339
0.1y - 0.09y = 1,339 - 1,260
0.01y = 79
0.01y/0.01 = 79/0.01
y = 7,900
Therefore, $7,900 was invested at the rate of 10% simple interest.
Let's determine x, substituting y = 7,900 in x + y = 14,000.
x + y = 14,000
x + 7,900 = 14,000
x = 14,000 - 7,900
x = 6,100
Therefore, $6,100 was invested at the rate of 9% simple interest.