A total of $13,000 is invested in two accounts. One of the two accounts pays 7% per year, and the other account pays 13% per year. If the total interest paid in the first year is $1,390, how much was invested in each account?
account paying 7% $
account paying 13% $
amount invested at 7% --- x
amount invested at 13% --- 13000-x
.07x + .13(13000-x) = 1390
solve for x
btw, it is college, not collage
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432
To solve this problem, we can use a system of equations. Let's assume that the amount invested in the account paying 7% is x dollars, and the amount invested in the account paying 13% is y dollars.
We can set up the following equations based on the given information:
1. The total amount invested is $13,000:
x + y = 13,000
2. The total interest earned in the first year is $1,390:
0.07x + 0.13y = 1,390
Now we have a system of two equations with two variables. We can solve this system to find the values of x and y.
There are various methods to solve this system, such as substitution or elimination. Let's use the elimination method.
First, we'll multiply both sides of equation 1 by 0.07 to make the coefficients of x in both equations the same:
0.07(x + y) = 0.07(13,000)
0.07x + 0.07y = 910
Now we'll subtract this equation from equation 2:
(0.07x + 0.13y) - (0.07x + 0.07y) = 1,390 - 910
0.13y - 0.07y = 480
0.06y = 480
y = 480 / 0.06
y = 8,000
Now that we have the value of y, we can substitute it back into one of the original equations (let's use equation 1) to find the value of x:
x + 8,000 = 13,000
x = 13,000 - 8,000
x = 5,000
Therefore, $5,000 was invested in the account paying 7% and $8,000 was invested in the account paying 13%.