How many solutions does the system of equations below have? 2x + 2y - 2z = -10 -x - 2y - 3z = 15 2x + 3y + 2z = -20 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Write Equations: First, let's write down the system of equations to see what we're working with: 1) 2x + 2y - 2z = -10 2) -x - 2y - 3z = 15 3) 2x + 3y + 2z = -20Eliminate Variable y: We can try to simplify the system by adding equations (1) and (2) to eliminate the variable y. So, (2x + 2y - 2z) + (-x - 2y - 3z) = -10 + 15 This simplifies to x - 5z = 5Correct Mistake: Now let's add equations (1) and (3) to eliminate y again. So, (2x + 2y - 2z) + (2x + 3y + 2z) = -10 - 20 This simplifies to 4x + 5y = -30 But wait, I made a mistake here. I should have canceled y, not kept it. Let's correct that.

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How many solutions does the system of equations below have? $\newline$ $2x + 2y - 2z = -10$ $\newline$ $-x - 2y - 3z = 15$ $\newline$ $2x + 3y + 2z = -20$ $\newline$ 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 - 2z = -10

\newline

-x - 2y - 3z = 15

\newline

2x + 3y + 2z = -20

\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 to see what we're working with: $\newline$ $1$ ) $2x + 2y - 2z = -10$ $\newline$ $2$ ) $-x - 2y - 3z = 15$ $\newline$ $3$ ) $2x + 3y + 2z = -20$
Eliminate Variable $y$ : We can try to simplify the system by adding equations ( $1$ ) and ( $2$ ) to eliminate the variable $y$ . So, $(2x + 2y - 2z) + (-x - 2y - 3z) = -10 + 15$ This simplifies to $x - 5z = 5$
Correct Mistake: Now let's add equations ( $1$ ) and ( $3$ ) to eliminate $y$ again. So, $(2x + 2y - 2z) + (2x + 3y + 2z) = -10 - 20$ This simplifies to $4x + 5y = -30$ But wait, I made a mistake here. I should have canceled $y$ , not kept it. Let's correct that.