Question

Linda invest $800 In one account and $1100 in an account paying 4% higher interest. At The end of one year she had earned $177 in interest. At what rates did she invest?

Answer 1

Let the rate for the $800 investment be q and

let the rate for the $1100 be r.

Therefore,

`r=q+0.04------------(1)`
`800q+1100r=117----------(2)`

Substituting equation (1) into equation (2), we have

`\begin{gathered} 800q+1100(q+0.04)=117 \\ 800q+1100q+44=117 \\ 1900q=73 \\ \text{ Hence} \\ q=(73)/(1900)\approx0.038 \\ \text{ therefore,} \\ r=0.038+0.04 \\ r=0.078 \end{gathered}`

Hence the rate for the $800 investment is 3.8% and

the rate for the $1100 investment is 7.8%