Question

Determine if the given ordered triple is a solution of the system. (2,−1,3)(2,−1, 3)  ﻿{x+y+z=4x−2y−z=12x−y−2z=−1\begin{cases}
x + y + z = 4 \
x - 2y - z = 1 \
2x - y - 2z = -1
\end{cases}⎩⎨⎧​x+y+z=4x−2y−z=12x−y−2z=−1​

Accepted Answer

Identify the system of equations and the ordered triple to test. The system is:

$$x + y + 0z = 4$$
$$x - 2y - 0z = 1$$
$$2x - y - 2z = -1$$

and the ordered triple is $$(2, -1, 3)$$, where $$x=2$$, $$y=-1$$, and $$z=3. $$Substitute the values of $$x$$, $$y$$, and $$z$$ from the ordered triple into the first equation:

$$2 + (-1) + 0 	imes 3 = ?$$

Simplify the left side to check if it equals 4. Substitute the values of $$x$$, $$y$$, and $$z$$ into the second equation:

$$2 - 2 	imes (-1) - 0 	imes 3 = ?$$

Simplify the left side to check if it equals 1. Substitute the values of $$x$$, $$y$$, and $$z$$ into the third equation:

$$2 	imes 2 - (-1) - 2 	imes 3 = ?$$

Simplify the left side to check if it equals -1. If all three simplified expressions equal their respective right-hand side values, then the ordered triple is a solution to the system. Otherwise, it is not.

Determine if the given ordered triple is a solution of the system. $open paren 2 comma negative 1 comma 3 close paren$
$\begin{cases}\)x + y + z = 4 \(\x\) - 2y - z = 1 \\2x - y - 2z = -1\(\end{cases}$

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Key Concepts

Ordered Triples as Solutions

Systems of Linear Equations

Substitution Method

Determine if the given ordered triple is a solution of the system. (2,−1,3)(2,−1, 3) {x+y+z=4x−2y−z=12x−y−2z=−1\(\begin{cases}\)x + y + z = 4 \(\x\) - 2y - z = 1 \\2x - y - 2z = -1\(\end{cases}\)⎩⎨⎧x+y+z=4x−2y−z=12x−y−2z=−1