How many solutions does the system of equations below have? -x - 2y - z = 14 -2x + y + 3z = 18 2x - y - 3z = -17 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Write Equations: First, let's write down the system of equations: 1) -x - 2y - z = 14 2) -2x + y + 3z = 18 3) 2x - y - 3z = -17Elimination Method: We can try to solve the system using the elimination or substitution method. Let's start by adding equations 2) and 3) to eliminate y and z. (-2x + y + 3z) + (2x - y - 3z) = 18 + (-17)Combine Equations: Simplifying the left side, we get: -2x + 2x + y - y + 3z - 3z = 0Simplify Left Side: Simplifying the right side, we get: 18 - 17 = 1Simplify Right Side: But on the left side, all the variables cancel out, so we are left with: 0 = 1 This is a contradiction, which means there is no solution to the system of equations.

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How many solutions does the system of equations below have? $\newline$ $-x - 2y - z = 14$ $\newline$ $-2x + y + 3z = 18$ $\newline$ $2x - y - 3z = -17$ $\newline$ Choices: $\newline$ (A)no solution $\newline$ (B)one solution $\newline$ (C)infinitely many solutions

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Precalculus

Determine the number of solutions to a system of equations in three variables

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Q. How many solutions does the system of equations below have?

\newline

-x - 2y - z = 14

\newline

-2x + y + 3z = 18

\newline

2x - y - 3z = -17

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Write Equations: First, let's write down the system of equations: $\newline$ $1$ ) $-x - 2y - z = 14$ $\newline$ $2$ ) $-2x + y + 3z = 18$ $\newline$ $3$ ) $2x - y - 3z = -17$
Elimination Method: We can try to solve the system using the elimination or substitution method. Let's start by adding equations $2)$ and $3)$ to eliminate $y$ and $z$ . $(-2x + y + 3z) + (2x - y - 3z) = 18 + (-17)$
Combine Equations: Simplifying the left side, we get: $-2x + 2x + y - y + 3z - 3z = 0$
Simplify Left Side: Simplifying the right side, we get: $18 - 17 = 1$
Simplify Right Side: But on the left side, all the variables cancel out, so we are left with: $\newline$ $0 = 1$ $\newline$ This is a contradiction, which means there is no solution to the system of equations.