We can solve this problem using a system of linear equations, have in mind the expression for simple interest:
[tex]\begin{gathered} I=\text{prt} \\ I=\text{interest} \\ p=\text{principal} \\ r=\text{rate in decimal form} \\ t=\text{time in years} \end{gathered}[/tex]Let x be the principal invested on the first account.
Let y be the principal invested on the second account.
[tex]\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}[/tex]We can solve the system by the method of substitution, isolate the variable y in equation (1):
[tex]y=30,000-x\text{ (1)}[/tex]Now, substitute (1) into equation (2) to get x-value:
[tex]\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}[/tex]Austin invested $16,000 on the first account.
Then, substitute the x-value in the equation (1) to get y-value:
[tex]\begin{gathered} y=30,000-16,000 \\ y=14,000 \end{gathered}[/tex]Austin invested $14,000 on the second account.