How many solutions does the system of equations below have? 3x + y + 2z = -15 -3x - y - 2z = 15 -2x - 2y + z = 12 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Addition of Equations: Add the first two equations to see if they cancel each other out. 3x + y + 2z = -15 -3x - y - 2z = 15 ----------------- 0 = 0Verification of Equations: Since 0 = 0 is a true statement, it means the first two equations are multiples of each other and represent the same plane.Checking Third Equation: Now, check if the third equation is also a multiple of the first one. -2x - 2y + z = 12 If we multiply the first equation by -2/3, we should get the third equation if they are the same plane. (-2/3)(3x + y + 2z) = (-2/3)(-15) -2x - (2/3)y - (4/3)z = 10 This is not the same as the third equation, so there is a mistake here.

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How many solutions does the system of equations below have? $\newline$ $3x + y + 2z = -15$ $\newline$ $-3x - y - 2z = 15$ $\newline$ $-2x - 2y + z = 12$ $\newline$ 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

3x + y + 2z = -15

\newline

-3x - y - 2z = 15

\newline

-2x - 2y + z = 12

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Addition of Equations: Add the first two equations to see if they cancel each other out. $\newline$ $3x + y + 2z = -15$ $\newline$ $-3x - y - 2z = 15$ $\newline$ ----------------- $\newline$ $0 = 0$
Verification of Equations: Since $0 = 0$ is a true statement, it means the first two equations are multiples of each other and represent the same plane.
Checking Third Equation: Now, check if the third equation is also a multiple of the first one. $\newline$ $-2x - 2y + z = 12$ $\newline$ If we multiply the first equation by $-2/3$ , we should get the third equation if they are the same plane. $\newline$ $(-2/3)(3x + y + 2z) = (-2/3)(-15)$ $\newline$ $-2x - (2/3)y - (4/3)z = 10$ $\newline$ This is not the same as the third equation, so there is a mistake here.