Answer
$\$2600$ invested at $12\%$
$\$2600$ invested at $14\%$
Work Step by Step
Let's note:
$x$=the amount invested at $12\%$
$y$=the amount invested at $14\%$
We can write the system:
$\begin{cases}
x+y=4000\\
0.12x+0.14y=508
\end{cases}$
We will use the addition method. Multiply Equation 1 by -0.14 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
-0.14(x+y)=-0.14(4000)\\
0.12x+0.14y=508
\end{cases}$
$\begin{cases}
-0.14x-0.14y=-560\\
0.12x+0.14y=508
\end{cases}$
$-0.14x-0.14y+0.12x+0.14y=-560+508$
$-0.02x=-52$
$x=\dfrac{-52}{-0.02}$
$x=2600$
$x+y=4000$
$2600+y=4000$
$y=4000-2600$
$y=1400$
The system's solution is:
$x=2600$
$y=1400$
