Mary has a total of $5000 invested in two accounts. One account pays 5% and the other 8%. Her interest in the first year was $331. Write and solve a system of equations to find out how much she has invested in both accounts.
amount invested at 5% --- x
amount invested at 8% --- y
x+y = 5000
(5/100)x + (8/100)y = 331
times 100
5x + 8y = 33100
first equation times 5
5x + 5y = 25000
subtract them:
3y = 8100
y = 2700
take over....
2300
To solve this problem, we can set up a system of equations based on the given information. Let's assume Mary invested x dollars in the account paying 5% interest and y dollars in the account paying 8% interest.
The first equation represents the total amount invested:
x + y = 5000 ------------------- Equation 1
The second equation represents the total interest earned in the first year:
0.05x + 0.08y = 331 ------- Equation 2
Now, we can solve this system of equations.
We can begin by solving Equation 1 for x:
x = 5000 - y
Next, substitute this value of x into Equation 2:
0.05(5000 - y) + 0.08y = 331
Now, simplify the equation:
250 - 0.05y + 0.08y = 331
250 + 0.03y = 331
0.03y = 331 - 250
0.03y = 81
Finally, solve for y:
y = 81 / 0.03
y = 2700
Substitute the value of y back into Equation 1 to find the value of x:
x + 2700 = 5000
x = 5000 - 2700
x = 2300
So, Mary has invested $2300 in the account paying 5% interest and $2700 in the account paying 8% interest.