An investment adviser invested $14,000 in two accounts. One investment earned 4% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of interest earned for 1 year was $458. How much was invested in each account
x and 14000-x
.04 x + .025 (14,000-x) = 458
solve for x and then do 14,000-x
To solve this problem, we can set up a system of equations based on the given information.
Let's assume that the amount invested at 4% interest is x dollars, and the amount invested at 2.5% interest is y dollars.
According to the problem, the total amount invested is $14,000, so we have the equation:
x + y = 14,000
The interest earned from the 4% account is calculated as 4% of x, which is 0.04x. Similarly, the interest earned from the 2.5% account is calculated as 2.5% of y, which is 0.025y. A combined total of $458 in interest was earned, so we have the equation:
0.04x + 0.025y = 458
Now, we can solve this system of equations using either substitution or elimination method. Let's solve it using substitution:
From the first equation, we have x = 14,000 - y.
Substituting this into the second equation, we get:
0.04(14,000 - y) + 0.025y = 458
Distributing the 0.04:
560 - 0.04y + 0.025y = 458
Combining like terms:
0.015y = 458 - 560
0.015y = -102
Dividing both sides by 0.015, we find:
y = -102 / 0.015
y = -6800
This means that the amount invested in the 2.5% account is $6,800.
To find the amount invested in the 4% account, we can substitute this value of y back into the first equation:
x + 6,800 = 14,000
Subtracting 6,800 from both sides, we get:
x = 14,000 - 6,800
x = 7,200
So, the amount invested in the 4% account is $7,200.
To summarize, $7,200 was invested in the account with a 4% interest rate, and $6,800 was invested in the account with a 2.5% interest rate.