# Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 831: 39

The amounts invested (in dollars): $1200$ ($8\%$), $2000$ ($10\%$), $3500$ ($12\%$)

#### Work Step by Step

Step 1. Assume the amounts invested (in dollars) are: $x$ ($8\%$), $y$ ($10\%$), $z$ ($12\%$), Step 2. Based on the given conditions, we have $\begin{cases} x+y+z=6700\\ 0.08x+0.1y+0.12z=716 \\ z=x+y+300 \end{cases}$ Step 3. Simplify and rearrange the above to get $\begin{cases} x+y+z=6700\\ 4x+5y+6z=35800 \\ x+y-z=-300 \end{cases}$ Step 4. Eliminate $x+y$ using equations 1 and 3; we have $2z=7000$, thus $z=3500$ Step 5. Use the z-value; we get $\begin{cases} x+y=3200\\ 4x+5y=14800 \end{cases}$ Step 6. Multiply 4 to the first equation and take the difference from the second; we get $y=2000$. Thus $x=1200$ Step 7. We conclude that the amounts invested (in dollars) are: $1200$ ($8\%$), $2000$ ($10\%$), $3500$ ($12\%$),

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