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

View step-by-step help

Home

Math Problems

Precalculus

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

Full solution

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.