Question

Sam initially invested $4,500 into a savings account that offers an interest rate of 3% each year. He wants to determine the number of years, t, for which the account will have less than or equal to $7,020. Create the inequality that represents this situation and plot the solution set to the inequality on the number line.

Answer 1

`\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&7020\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 3\%\to (3)/(100)\dotfill &0.03\\ t=years\dotfill \end{cases}`

`\bf 7020\geqslant 4500(1+0.03t)\implies \cfrac{7020}{4500}\geqslant 1+0.03t\implies \cfrac{39}{25}\geqslant 1+0.03t \\\\\\ \cfrac{39}{25}-1\geqslant 0.03t\implies \cfrac{14}{25}\geqslant 0.03t\implies \cfrac{~~(14)/(25)~~}{0.03}\geqslant t\implies \cfrac{56}{3}\geqslant t \\\\\\ 18(2)/(3)\geqslant t\impliedby \textit{18 years and 8 months}~\hfill \boxed{0 \leqslant t \leqslant 18(2)/(3)} \\\\\\ ~\hspace{34em}`

check the picture below.

bearing in mind that "t" cannot be less than 0.