Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 52

Piotr Galkowski invested some money at $4.5\%$ simple interest and $\$1,000$ less
than twice this amount at $3\%$. Her total annual income from the interest was $\$1,020$. How much was invested at each rate?

Step 1: Read the problem, we are asked to find the amount invested at each rate.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x =$ amount invested in $3.5\%$ interest rate
Then, $3x + 5,000 =$ amount invested in $4\%$ interest rate

$\begin{array}{|c|c|c|c|c|c|} \hline & \rm{Principal} & \cdot & \text{Interest Rate} & = & \rm{Interest} \\ \hline 3.5\% & x & \cdot & 0.035 & = & 0.035x \\ \hline 4\% & 3x + 5,000 & \cdot & 0.04 & = & 0.04(3x + 5,000) \\ \hline \end{array}$



The total interest earned is equal to the sum of the interests at each rate.

Step 3: Write an equation from the last column of the table
$0.035x + 0.04(3x + 5,000) = 1,440$

Step 4: Solve

$\begin{equation} \begin{aligned} 0.035x + 0.04(3x + 5,000) &= 1,440\\ \\ 0.035x + 0.12x + 200 &= 1,440\\ \\ 0.155x &= 1,440 - 200 \\ \\ 0.155x &= 1,240\\ \\ x &= 8,000 \end{aligned} \end{equation}$


Then by substitution,
$3x + 5,000 = 3(8,000) + 5,000 = 29,000$

Step 5: State the answer
In other words, the amount invested in $3.5\%$ and $4\%$ interest rate is $\$8,000$ and $\$29,000$ respectively.