mrs. ming invested an amount of money in two accounts for one year. she invested some at 8% interest and the…

mrs. ming invested an amount of money in two accounts for one year. she invested some at 8% interest and the rest at 6% interest. her total amount invested was $1,500. at the end of the year, she had earned $106.40 in interest. how much had mrs. ming invested in the account paying 6%?\n$117\n$680\n$760\n$820

mrs. ming invested an amount of money in two accounts for one year. she invested some at 8% interest and the rest at 6% interest. her total amount invested was $1,500. at the end of the year, she had earned $106.40 in interest. how much had mrs. ming invested in the account paying 6%?\n$117\n$680\n$760\n$820

Answer

Explanation:

Step1: Let the amount invested at 6% be $x$.

Then the amount invested at 8% is $1500 - x$.

Step2: Calculate the interest from each account.

The interest from the 6% - account is $0.06x$, and the interest from the 8% - account is $0.08(1500 - x)$.

Step3: Set up the interest - equation.

The total interest is $106.40$, so $0.06x+0.08(1500 - x)=106.40$.

Step4: Expand and simplify the equation.

$0.06x + 120-0.08x=106.40$. Combine like - terms: $120-(0.08x - 0.06x)=106.40$, which gives $120 - 0.02x=106.40$.

Step5: Solve for $x$.

Subtract 120 from both sides: $-0.02x=106.40 - 120=-13.60$. Then divide both sides by $- 0.02$: $x=\frac{-13.60}{-0.02}=680$.

Answer:

$680$