Example 3 from Algebra and Trigonometry, 11.2 Systems of Linear Equations: Three Variables

Solving a Real-World Problem Using a System of Three Equations in Three Variables

In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:

$\begin{array}{l} x = amount invested in money-market fund \\ y = amount invested in municipal bonds \\ z = amount invested in mutual funds \end{array}$
The first equation indicates that the sum of the three principal amounts is $12,000.

$x + y + z = 12,000$
We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.

$z = y + 4,000$
The third equation shows that the total amount of interest earned from each fund equals $670.

$0.03 x + 0.04 y + 0.07 z = 670$

Then, we write the three equations as a system.

$\begin{array}{l} x+y+z=12,000 \\ -y+z=4,000 \\ 0.03 x + 0.04 y + 0.07 z = 670 \end{array}$
To make the calculations simpler, we can multiply the third equation by 100. Thus,

$\begin{array}{l} x+ & (1) \\ -y+z & (2) \\ 3 x + 4 y + 7 z = 67,000 & (3) \end{array}$
Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

$\begin{array}{l} x+ \\ 3 x + 4 y + 7 z = 67,000 \\ -y \end{array}$
Step 2. Multiply equation (1) by $−3$ and add to equation (2). Write the result as row 2.
$\begin{array}{l} x + y + z =12,000 \\ y + 4 z = 31,000 \\ -y+z \end{array}$
Step 3. Add equation (2) to equation (3) and write the result as equation (3).

$\begin{array}{l} x + y + z=12,000 \\ y+4z=31,000 \\ 5z \end{array}$
Step 4. Solve for $z$ in equation (3). Back-substitute that value in equation (2) and solve for $y .$ Then, back-substitute the values for $z$ and $y$ into equation (1) and solve for $x .$

$\begin{array}{l} 5z=35,000 \\ z=7,000 \\ y+4(7,000)=31,000 \\ y=3,000 \\ x + 3,000 + 7,000 = 12,000 \\ x=2,000 \end{array}$
John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system of equations in three variables.

\begin{array}{l} 2 x + y −2 z = −1 \\ 3 x −3 y - z = 5 \\ x −2 y + 3 z = 6 \end{array}

How did it go?

Solution

Your turn

Solution

Your turn