Question

Last year, Deon had $20,000 to invest. He invested some of it in an account that paid 7% simple interest per year, and he invested the rest in an account thatpaid 8% simple interest per year. After one year, he received a total of $1420 in interest. How much did he invest in each account?

Answer 1

Let's call the amount Deon invested on the 7% account as x, and the amount he invested on the 8% account as y. Since Deon had $20,000 to invest, the sum of those amounts must add up to 20,000.

`x+y=20000`

The interest of a simple interest account is given by the following formula

`I=P* r* t`

Where P represents the principal(starting value), r represents the interest rate(in decimals) and t represents the amount of time.

Both accounts received interest for 1 year, therefore, t = 1. Now, the interest on each account will be different because the amount invested is different and the rate is different. For the first account, the interest is given by

`I_x=x*0.07*1=0.07x`

The interest on the second account will be

`I_y=y*0.08*1=0.08y`

And the total interest will be the sum of those values.

`I=I_x+I_y=0.07x+0.08y=1420`

Combining this equation with the first equation for the distinct amounts, we have a linear system.

`\begin{cases}x+y=20000 \\ 0.07x+0.08y=1420\end{cases}`

If we rewrite the first equation with x as a function of y

`x+y=20000\Rightarrow x=20000-y`

and substitute this on the second expression

`\begin{gathered} 0.07x+0.08y=1420 \\ 0.07(20000-y)+0.08y=1420 \\ 1400-0.07y+0.08y=1420 \\ 1400+0.01y=1420 \end{gathered}`

We have an expression only for y. Solving for y, we have

`\begin{gathered} 1400+0.01y=1420 \\ 1400+0.01y-1400=1420-1400 \\ 0.01y=20 \\ y=(20)/(0.01) \\ y=2000 \end{gathered}`

Using this y value on any of the previous expressions, we get our x value.

`\begin{gathered} x+y=20000 \\ x+(2000)=20000 \\ x+2000-2000=20000-2000 \\ x=18000 \end{gathered}`

$18,000 were investe in an account that paid 7% simple interest per year, and $2,000 in an account that paid 8% simple interest per year.