12000 was invested in 2 accounts, one earning 12% simple interest and the other earning 8% simple interest. if the total interest at the end of 1yr was $1240, how much was invested in each account?
x = amount at 12%
y = amount at 8%
x + y = 12000
.12x +.08y = 1240
multiply by 100 to clear decimals
12x + 8y = 124000
-8x -8y = -96000
4x = 28000
x = 7000
y = 5000
To solve this problem, we can set up a system of equations.
Let's say x is the amount invested in the account earning 12% simple interest, and y is the amount invested in the account earning 8% simple interest.
We know that the total amount invested is $12,000, so we have the equation:
x + y = 12,000 -- Equation 1
We also know that at the end of 1 year, the total interest earned was $1,240. The interest earned from the account earning 12% is 12% of x, which is 0.12x, and the interest earned from the account earning 8% is 8% of y, which is 0.08y. Therefore, we have the equation:
0.12x + 0.08y = 1,240 -- Equation 2
Now, we can solve the system of equations using the substitution or elimination method.
Let's solve using the substitution method:
Solve Equation 1 for x:
x = 12,000 - y
Substitute this expression for x in Equation 2:
0.12(12,000 - y) + 0.08y = 1,240
Expand and simplify the equation:
1,440 - 0.12y + 0.08y = 1,240
1,440 - 0.04y = 1,240
Subtract 1,440 from both sides of the equation:
-0.04y = -200
Divide both sides of the equation by -0.04:
y = -200 / -0.04
y = 5,000
Now substitute the value of y back into Equation 1 to find x:
x + 5,000 = 12,000
x = 12,000 - 5,000
x = 7,000
Therefore, $7,000 was invested in the account earning 12% simple interest, and $5,000 was invested in the account earning 8% simple interest.