Chelsea invested her savings of $4800. She invested part in a mutual fund that paid 9% interest per year, and the rest went into an account for 10% interest per year. After 1 year, the interest from the mutual fund was $43 less than the interest she earned from the other account. How much did Chelsea invest into each account?
.09x = .10(4800-x) - 43
amount invested in mutual fund at 9% : x
amount invested at 10% : 4800-x
interest from the mutual fund = .09x
interest from other account = .10(4800-x)
.10(4800-x) - .09x = 43
multiply by 100
100(4800-x) - 9x = 4300
solve for x
To solve this problem, we need to set up equations based on the given information.
Let's say Chelsea invested an amount, x, in the mutual fund that pays 9% interest per year. Therefore, the amount she invested in the other account that pays 10% interest per year would be (4800 - x).
The interest earned from the mutual fund can be calculated as (x * 9/100), and the interest earned from the other account is ((4800 - x) * 10/100).
From the given information, we know that the interest earned from the mutual fund is $43 less than the interest earned from the other account. So we have the equation:
(x * 9/100) = ((4800 - x) * 10/100) + 43
Now, let's solve this equation to find the value of x, which represents the amount Chelsea invested in the mutual fund.
Multiply the equation by 100 to eliminate the fractions:
9x = 10(4800 - x) + 4300
Expand the equation:
9x = 48000 - 10x + 4300
Combine like terms:
19x = 52300
Divide both sides of the equation by 19:
x = 52300/19
x ≈ 2752.63
So Chelsea invested approximately $2752.63 in the mutual fund (at 9% interest), and the remaining amount would be approximately (4800 - 2752.63) ≈ $2047.37, which she invested in the other account (at 10% interest).