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# Linear Systems in Three Variables for Intermediate Algebra

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By — McGraw-Hill Professional
Updated on Aug 12, 2011

Practice problems for these concepts can be found at:

A solution of an equation in three variables, such as x – 2y + 3z = –4, is an ordered triple of real numbers (x, y, z). Values for each of the three variables must be substituted into the equation before we can decide whether the result is a true statement. The values of the components in the ordered triple are listed in alphabetical order. Thus, (3, 2, –1) and (–4, 0, 0) are solutions of the equation, while (1,0,1) is not. There are infinitely many ordered triples in this solution set.

The solution set of a system of three linear equations in three unknowns, such as

is the intersection of the solution sets of the individual equations in the system.

The system is a 3 × 3 (read 3 by 3) system since it consists of three equations in three unknowns. The graph of a first-degree equation in two variables is a straight line in a two-dimensional coordinate system. The graph of a first-degree equation in three variables is a plane in a three-dimensional coordinate system. Each equation in a 3 × 3 system therefore represents a plane in three-dimensional space. Three planes may intersect at a unique point, may intersect at infinitely many points, or may not intersect at any points common to all three planes. There are therefore three possibilities for the solution set of a 3 × 3 system;

1. There is a unique solution of the system. It is consistent.
2. There are infinitely many solutions of the system. It is dependent.
3. There is no solution of the system. It is inconsistent.

We employ methods similar to those used to solve systems of linear equations in two variables to solve a 3 × 3 system. We eliminate variables from pairs of equations until an equation in one unknown is obtained. We then employ "back substitution" to find remaining unknowns. We illustrate by solving the system given above.

Look at the system and choose a variable to eliminate first. Sometimes one variable can be eliminated more easily than the others. We shall eliminate y first using equations (1) and (3).

Now choose two different equations and eliminate y again. We shall use (2) and (3).

We now form a 2 × 2 system that consists of equations (4) and (5). Use the methods of the previous section to solve the system.

"Back substitute" by substituting x = 3 into (4) or (5) to find z = –1. Now substitute x = 3 and z = –1 into one of the original equations to find y. We choose equation (2) and find y = 2.

The solution is x = 3, y = 2, and z = –1 or (3, 2, –1). The solution set is {(3, 2, –1)}. Check: Verify that (3, 2, –1) satisfies each of the original equations.

The procedure used is summarized below. To solve a linear system in three variables:

1. Eliminate any variable from any pair of equations. Try to choose the most convenient variable to eliminate if there is one.
2. Use appropriate steps to eliminate the same variable as in step 1 from a different pair of the original equations.
3. Solve the resulting system of two equations in two variables.
4. Substitute the values obtained in step 3 into one of the original equations. Solve for the remaining variable.
5. Check the solution in each of the original equations.

If the resulting equation vanishes or yields a contradiction at any step in the process, the system contains dependent equations or else two or three inconsistent equations. The system then has infinitely many solutions or no solution, respectively.

See solved problem 7.4.

Practice problems for these concepts can be found at:

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