Answer
$x=1200$
$y=2000$
$z=3500$
Work Step by Step
We denote the amount of of money invested in $8\%$ rate by $x$, the amount of money invested in $10\%$ rate by $y$ and the amount of money invested in $12\%$ rate by $z$.
We are given the system:
$\begin{cases}
x+y+z=6700\\
0.08x+0.10y+0.12z=716\\
z=x+y+300
\end{cases}$
We will use the addition method. first substract (x+y) from both sides of the equation to get the variable on one side in Equation 3. Add Equation 3 to Equation 1 to eliminate $x+y$.
$\begin{cases}
z-x-y+x+y+z=300+6700,\\
2z=7000,\\
z=3500,\\
\end{cases}$
substitute $z$ result we got into Equation 1 and Equation 2. Multiply Equation 2 by -10 and it to Equation 1 to eliminate $y$.
$\begin{cases}
x+y-0.8x-y=3200-2960,\\
0.2x=240,\\
x=1200
\end{cases}$
substitute the result we got for $x$ and $z$ into Equation 1 to solve for $y$.
$1200+y+3500=6700,
y=2000$
The solution to the systems are:
$x=1200$
$y=2000$
$z=3500$
