Question

George invested a total of $5,000 at the beginning of the year in two different funds. At the end of the year, his investment had grown to $5,531. The money in the first fund earned 9%, while the money in the second fund earned 13.5%. Write a system of equations, then solve it to find out how much of the $5,000 was invested into each fund at the beginning of the year

Answer 1

The amount invested at 9% was $3,200 and the amount invested at 13.5% was $1,800

Explanation:

Let

x ----> the amount invested at 9% (first fund)

5,000-x ----> the amount invested at 13.5% (second fund)

Remember that

`9\%=9/100=0.09`

`13.5\%=13.5/100=0.135`

The total interest earned is equal to

`\$5,531-\$5,000=\$531`

we know that

The amount earned by the first fund at 9% plus the amount earned by the second fund at 13.5% must be equal to $531

so

the linear equation that represent this situation is equal to

`0.09x+0.135(5,000-x)=531`

solve for x

`0.09x+675-0.135x=531`

`0.135x-0.09x=675-531`

`0.045x=144`

`x=\$3,200`

so

`\$5,000-x=\$5,000-\$3,200=\$1,800`

therefore

The amount invested at 9% was $3,200 and the amount invested at 13.5% was $1,800