Solve Systems of Linear Equations in Three Variables
-$\begin{array} { r r } x + y & = - 2 \ 3 x + 5 y + 3 z & = - 18 \ x - z & = 3 \end{array}$
A) $\{ ( 1 , - 3 , - 2 ) \}$
B) $\{ ( 1 , - 2 , - 3 ) \}$
C) $\{ ( - 2 , - 3,1 ) \}$
D) $\{ ( - 2,1 , - 3 ) \}$

Question

Quizplus · Accepted Answer

To solve the system of equations, we can use substitution or elimination. From the third equation, $x = z + 3$. Substituting $x$ from the third equation into the first and second equations:1. Substituting $x = z + 3$ into $x + y = -2$, we get $z + 3 + y = -2$, which simplifies to $y = -z - 5$.2. Substituting $x = z + 3$ into $3x + 5y + 3z = -18$, we get $3(z + 3) + 5y + 3z = -18$. Using $y = -z - 5$, we can solve for $z$.After substitution, we find that $z = -2$, $x = 1$ (from $x = z + 3$), and $y = -3$ (from $y = -z - 5$). Therefore, the solution is $\{(1, -3, -2)\}$.

Solve Systems of Linear Equations in Three Variables - x+y=−23x+5y+3z=−18x−z=3\begin{array} { r r } x + y & = - 2 \\ 3 x + 5 y + 3 z & = - 18 \\ x - z & = 3 \end{array}x+y3x+5y+3zx−z​=−2=−18=3​

Solve Systems of Linear Equations in Three Variables
- $\begin{array} { r r } x + y & = - 2 \\ 3 x + 5 y + 3 z & = - 18 \\ x - z & = 3 \end{array}$