Question

last year, Austin had $30,000 to invest he invested some of it an account that paid 8% simple interest per year,and she invest the rest in a bank account that paid 10% simple interest per year.after one year, he receives a total of $2,680 to invest.how much did he and invest in each account?first account:____$second account:___$

Answer 1

We can solve this problem using a system of linear equations, have in mind the expression for simple interest:

$\begin{gathered} I=\text{prt} \\ I=\text{interest} \\ p=\text{principal} \\ r=\text{rate in decimal form} \\ t=\text{time in years} \end{gathered}$

Let x be the principal invested on the first account.

Let y be the principal invested on the second account.

$\begin{gathered} x+y=30,000\rightarrow\operatorname{Re}present\text{ total invested (1)} \\ 0.08x+0.1y=2,680\rightarrow\operatorname{Re}present\text{ the rates and interest earned }(2) \end{gathered}$

We can solve the system by the method of substitution, isolate the variable y in equation (1):

$y=30,000-x\text{ (1)}$

Now, substitute (1) into equation (2) to get x-value:

$\begin{gathered} 0.08x+0.1(30,000-x)=2,680 \\ 0.08x+3,000-0.1x=2,680 \\ -0.02x=2,680-3,000 \\ -0.02x=-320 \\ x=\frac{320}{0.02} \\ x=16,000 \end{gathered}$

Austin invested $16,000 on the first account.

Then, substitute the x-value in the equation (1) to get y-value:

$\begin{gathered} y=30,000-16,000 \\ y=14,000 \end{gathered}$

Austin invested $14,000 on the second account.