How many solutions does the system of equations below have? 2x - 2y - z = 3 -x - y + z = -12 -3x + y + 3z = -12 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Multiply and Add Equations: First, let's multiply the second equation by 2 so we can add it to the first equation and eliminate y. 2(-x - y + z) = 2(-12) -2x - 2y + 2z = -24Combine Equations: Now, add the new equation to the first one: (2x - 2y - z) + (-2x - 2y + 2z) = 3 + (-24) The x terms and z terms cancel out, leaving: -4y = -21

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How many solutions does the system of equations below have? $\newline$ $2x - 2y - z = 3$ $\newline$ $-x - y + z = -12$ $\newline$ $-3x + y + 3z = -12$ $\newline$ 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

2x - 2y - z = 3

\newline

-x - y + z = -12

\newline

-3x + y + 3z = -12

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Multiply and Add Equations: First, let's multiply the second equation by $2$ so we can add it to the first equation and eliminate $y$ . $\newline$ $2(-x - y + z) = 2(-12)$ $\newline$ $-2x - 2y + 2z = -24$
Combine Equations: Now, add the new equation to the first one: $\newline$ $(2x - 2y - z) + (-2x - 2y + 2z) = 3 + (-24)$ $\newline$ The $x$ terms and $z$ terms cancel out, leaving: $\newline$ $-4y = -21$