Question

Ms. Walsh invested $18,000 in two accounts, one yielding 8% interest and the other yielding 10%. If she received a

total of $1,600 in interest at the end of the year, how much did she invest in each account?

The amount invested at 10% was$?

Answer 1

Explanation:

Let x be the amount Ms. Walsh invested in the account yielding 8% interest, and let y be the amount she invested in the account yielding 10% interest.

According to the problem, the total amount invested is $18,000, so we have:

x + y = 18,000

The interest earned from the account yielding 8% interest is 0.08x, and the interest earned from the account yielding 10% interest is 0.10y. The total interest earned is $1,600, so we have:

0.08x + 0.10y = 1,600

We now have two equations with two unknowns. We can solve for one of the unknowns and then use that value to find the other unknown.

Solving the first equation for y, we get:

y = 18,000 - x

Substituting this expression for y into the second equation, we get:

0.08x + 0.10(18,000 - x) = 1,600

Simplifying and solving for x, we get:

0.08x + 1,800 - 0.10x = 1,600

-0.02x = -200

x = 10,000

Therefore, Ms. Walsh invested $10,000 in the account yielding 8% interest. To find the amount she invested in the account yielding 10% interest, we can substitute x = 10,000 into the equation y = 18,000 - x:

y = 18,000 - 10,000

y = 8,000

Therefore, Ms. Walsh invested $8,000 in the account yielding 10% interest.