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Determining whether an ordered triple is a solution to a system

Determine whether the ordered triple ( 3 , −2 , 1 ) is a solution to the system.

       x + y + z = 2 6 x 4 y + 5 z = 31 5 x + 2 y + 2 z = 13

We will check each equation by substituting in the values of the ordered triple for x , y , and z .

x + y + z = 2 ( 3 ) + ( −2 ) + ( 1 ) = 2 True 6 x −4 y + 5 z = 31 6 ( 3 ) −4 ( −2 ) + 5 ( 1 ) = 31 18 + 8 + 5 = 31 True 5 x + 2 y + 2 z = 13 5 ( 3 ) + 2 ( −2 ) + 2 ( 1 ) = 13 15 −4 + 2 = 13 True

The ordered triple ( 3 , −2 , 1 ) is indeed a solution to the system.

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Given a linear system of three equations, solve for three unknowns.

  1. Pick any pair of equations and solve for one variable.
  2. Pick another pair of equations and solve for the same variable.
  3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
  4. Back-substitute known variables into any one of the original equations and solve for the missing variable.

Solving a system of three equations in three variables by elimination

Find a solution to the following system:

     x −2 y + 3 z = 9 (1)     x + 3 y z = −6 (2) 2 x −5 y + 5 z = 17 (3)

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate x by adding equations (1) and (2).

      x 2 y + 3 z = 9 (1)    x + 3 y z = −6  (2)              y + 2 z = 3     (3)

The second step is multiplying equation (1) by −2 and adding the result to equation (3). These two steps will eliminate the variable x .

−2 x + 4 y 6 z = −18 ( 1 ) multiplied by 2 2 x 5 y + 5 z = 17 ( 3 ) ____________________________________              y z = −1     ( 5 )

In equations (4) and (5), we have created a new two-by-two system. We can solve for z by adding the two equations.

y + 2 z = 3      ( 4 ) y z = 1    ( 5 ) z = 2     ( 6 )

Choosing one equation from each new system, we obtain the upper triangular form:

x −2 y + 3 z = 9   ( 1 )          y + 2 z = 3 ( 4 )                  z = 2 ( 6 )

Next, we back-substitute z = 2 into equation (4) and solve for y .

y + 2 ( 2 ) = 3       y + 4 = 3              y = −1

Finally, we can back-substitute z = 2 and y = −1 into equation (1). This will yield the solution for x .

x −2 ( −1 ) + 3 ( 2 ) = 9              x + 2 + 6 = 9                         x = 1

The solution is the ordered triple ( 1 , −1 , 2 ) . See [link] .

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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?

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:

x = amount invested in money-market fund y = amount invested in municipal bonds z = amount invested in mutual funds

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.

                     x + y + z = 12,000                        y + z = 4,000 0.03 x + 0.04 y + 0.07 z = 670

To make the calculations simpler, we can multiply the third equation by 100. Thus,

   x +     y + z   = 12,000 ( 1 )         y + z   = 4,000 ( 2 ) 3 x + 4 y + 7 z = 67,000 ( 3 )

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

   x +    y   +    z = 12,000 3 x + 4 y   + 7 z = 67,000          y    +    z = 4,000

Step 2. Multiply equation (1) by −3 and add to equation (2). Write the result as row 2.

x + y + z   = 12,000       y + 4 z = 31,000    y + z   = 4,000

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

x + y +    z = 12,000        y + 4 z = 31,000              5 z   = 35,000

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 .

                        5 z = 35,000                           z = 7,000         y + 4 ( 7,000 ) = 31,000                            y = 3,000   x + 3,000 + 7,000 = 12,000                                   x = 2,000

John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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