How many solutions does the system of equations below have? 3x + y + 3z = -5 2x - 3y - 3z = 16 3x + 2y + z = -17 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Solve System: First, let's try to solve the system using elimination or substitution.Eliminate z: We can multiply the second equation by 3 and add it to the first equation to eliminate z. (3x + y + 3z) + 3*(2x - 3y - 3z) = -5 + 3*16Combine Equations: This simplifies to 3x + y + 3z + 6x - 9y - 9z = -5 + 48.Add Third Equation: Combining like terms, we get 9x - 8y - 6z = 43.Check Consistency: Now, let's add the third equation to the new equation we just found. (9x - 8y - 6z) + (3x + 2y + z) = 43 - 17Form Coefficients Matrix: This simplifies to 12x - 6y - 5z = 26.Calculate Determinant: We can now try to solve for one of the variables, but we notice that we have three equations with three variables. We need to check if the equations are consistent and independent.Check Non-Zero Determinant: If we look at the coefficients of the variables in the original equations, we can form a matrix and check its determinant to see if the system has a unique solution.Unique Solution: The matrix formed by the coefficients is: | 3 1 3 | | 2 -3 -3 | | 3 2 1 |Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.The determinant is 3*(-3)*1 + 1*(-3)*3 + 3*2*2 - 3*(-3)*3 - 1*2*3 - 3*2*1.This simplifies to -9 + -9 + 12 - (-27) - 6 - 6.The determinant is -9 - 9 + 12 + 27 - 6 - 6, which equals -9.Since the determinant is non-zero, the system has a unique solution.

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How many solutions does the system of equations below have? $\newline$ $3x + y + 3z = -5$ $\newline$ $2x - 3y - 3z = 16$ $\newline$ $3x + 2y + z = -17$ $\newline$ 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

3x + y + 3z = -5

\newline

2x - 3y - 3z = 16

\newline

3x + 2y + z = -17

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Solve System: First, let's try to solve the system using elimination or substitution.
Eliminate z: We can multiply the second equation by $3$ and add it to the first equation to eliminate $z$ .
$(3x + y + 3z) + 3\times(2x - 3y - 3z) = -5 + 3\times16$
Combine Equations: This simplifies to $3x + y + 3z + 6x - 9y - 9z = -5 + 48$ .
Add Third Equation: Combining like terms, we get $9x - 8y - 6z = 43$ .
Check Consistency: Now, let's add the third equation to the new equation we just found. $\newline$ $(9x - 8y - 6z) + (3x + 2y + z) = 43 - 17$
Form Coefficients Matrix: This simplifies to $12x - 6y - 5z = 26$ .
Calculate Determinant: We can now try to solve for one of the variables, but we notice that we have $3$ equations with $3$ variables. We need to check if the equations are consistent and independent.
Check Non-Zero Determinant: If we look at the coefficients of the variables in the original equations, we can form a matrix and check its determinant to see if the system has a unique solution.
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$ Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$ Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.The determinant is $3(-3)1 + 1(-3)3 + 3\cdot2\cdot2 - 3(-3)3 - 1\cdot2\cdot3 - 3\cdot2\cdot1$ .
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$ Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.The determinant is $3(-3)1 + 1(-3)3 + 3\cdot2\cdot2 - 3(-3)3 - 1\cdot2\cdot3 - 3\cdot2\cdot1$ .This simplifies to $-9 + -9 + 12 - (-27) - 6 - 6$ .
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$ Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.The determinant is $3(-3)1 + 1(-3)3 + 3\cdot2\cdot2 - 3(-3)3 - 1\cdot2\cdot3 - 3\cdot2\cdot1$ .This simplifies to $-9 + -9 + 12 - (-27) - 6 - 6$ .The determinant is $-9 - 9 + 12 + 27 - 6 - 6$ , which equals $-9$ .
Unique Solution: The matrix formed by the coefficients is: $\newline$ $\begin{vmatrix} 3 & 1 & 3 \ 2 & -3 & -3 \ 3 & 2 & 1 \end{vmatrix}$ Calculating the determinant of this matrix, if it's non-zero, there is one solution. If it's zero, we have either no solution or infinitely many solutions.The determinant is $3(-3)1 + 1(-3)3 + 3\cdot2\cdot2 - 3(-3)3 - 1\cdot2\cdot3 - 3\cdot2\cdot1$ .This simplifies to $-9 + -9 + 12 - (-27) - 6 - 6$ .The determinant is $-9 - 9 + 12 + 27 - 6 - 6$ , which equals $-9$ .Since the determinant is non-zero, the system has a unique solution.