Difference between revisions of "Systems of Equations in Three Variables"
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* One solution: <math> x + y + z = 3 </math>, <math> 2x - 3y + z = 0 </math>, and <math> -5x - 5y + 23z = 13 </math>. <math> x + y + z = 3 \implies 5x + 5y + 5z = 15</math>. We can add this to the third equation to get <math> 28z = 28 </math>, which means z = 1. So, the first two equations can be rewritten as <math> x + y = 2 </math> and <math> 2x - 3y = -1 </math>. Using substitution, elimination, or graphing, we can calculate that x = 1 and y = 1 with these two equations. Thus, the solution to the system is (x, y, z) = (1, 1, 1). | * One solution: <math> x + y + z = 3 </math>, <math> 2x - 3y + z = 0 </math>, and <math> -5x - 5y + 23z = 13 </math>. <math> x + y + z = 3 \implies 5x + 5y + 5z = 15</math>. We can add this to the third equation to get <math> 28z = 28 </math>, which means z = 1. So, the first two equations can be rewritten as <math> x + y = 2 </math> and <math> 2x - 3y = -1 </math>. Using substitution, elimination, or graphing, we can calculate that x = 1 and y = 1 with these two equations. Thus, the solution to the system is (x, y, z) = (1, 1, 1). | ||
* No solutions: <math> x + y + z = 1 </math>, <math> x + y + z = 2 </math>, <math> x + y + z = 3 </math>. These equations represent three parallel planes, and there is no x, y, and z that satisfy all three equations simultaneously. So, this system has no solutions. | * No solutions: <math> x + y + z = 1 </math>, <math> x + y + z = 2 </math>, <math> x + y + z = 3 </math>. These equations represent three parallel planes, and there is no x, y, and z that satisfy all three equations simultaneously. So, this system has no solutions. | ||
| − | * Infinite solutions: <math> x + y = z </math>, <math> x + y = 2z </math>, <math> x + y = 4z </math>. x + y = 0 for all x and y such that y = -x. Since <math> z = 2z = 4z </math> when z = 0, this system | + | * Infinite solutions: <math> x + y = z </math>, <math> x + y = 2z </math>, <math> x + y = 4z </math>. x + y = 0 for all x and y such that y = -x. Since <math> z = 2z = 4z </math> when z = 0, this system has an infinite number of solutions of the form (x, -x, 0) where x can be any real number. (for example, <math> (3, -3, 0), (-1/2, 1/2, 0), (\pi, -\pi, 0)</math>, etc.). |
==Resources== | ==Resources== | ||
* [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math) | * [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math) | ||
Revision as of 10:53, 15 September 2021
See Systems of Equations in Two Variables for more information on systems of equations.
Examples
- One solution: , , and . . We can add this to the third equation to get , which means z = 1. So, the first two equations can be rewritten as and . Using substitution, elimination, or graphing, we can calculate that x = 1 and y = 1 with these two equations. Thus, the solution to the system is (x, y, z) = (1, 1, 1).
- No solutions: , , . These equations represent three parallel planes, and there is no x, y, and z that satisfy all three equations simultaneously. So, this system has no solutions.
- Infinite solutions: , , . x + y = 0 for all x and y such that y = -x. Since when z = 0, this system has an infinite number of solutions of the form (x, -x, 0) where x can be any real number. (for example, , etc.).
Resources
- Linear Systems with Three Variables, Paul's Online Notes (Lamar Math)
