Difference between revisions of "Systems of Equations in Three Variables"

Revision as of 10:38, 15 September 2021

See Systems of Equations in Two Variables for more information on systems of equations.

One solution: $x+y+z=3$ , $2x-3y+z=0$ , and $-5x-5y+23z=13$ . $x+y+z=3\implies 5x+5y+5z=15$ . We can add this to the third equation to get $28z=28$ , which means z = 1. So, the first two equations can be rewritten as $x+y=2$ and $2x-3y=-1$ . 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).

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 See [[Systems of Equations in Two Variables]] for more information on systems of equations.
 ==Examples==
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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).
 ==Resources==
 * [https://tutorial.math.lamar.edu/classes/alg/systemsthreevrble.aspx Linear Systems with Three Variables], Paul's Online Notes (Lamar Math)