Solve the system. If there is no solution or if the system's equations are dependent, so state.
-$$\left\{ \begin{array} { r r }
x - y + 3 z & = - 1 \
4 x + z & = 0 \
x + 5 y + z & = 5
\end{array} \right.$$
A) infinitely many solutions; dependent equations
B) no solution or $\emptyset \varnothing$
C) $\{ ( 0,1,0 ) \}$
D) $\{ ( 0,0,1 ) \}$

Question

Solve the system. If there is no solution or if the system's equations are dependent, so state.
-$$\left\{ \begin{array} { r r } 
x - y + 3 z & = - 1 \
4 x + z & = 0 \
x + 5 y + z & = 5
\end{array} \right.$$
A) infinitely many solutions; dependent equations
B) no solution or $\emptyset \varnothing$
C) $\{ ( 0,1,0 ) \}$
D) $\{ ( 0,0,1 ) \}$

Quizplus · Accepted Answer

To solve the system, we can use substitution or elimination. From the second equation, $4x + z = 0$, we can express $z$ in terms of $x$: $z = -4x$. Substituting $z = -4x$ into the first and third equations, we get:1) $x - y - 12x = -1$ which simplifies to $-11x - y = -1$,2) $x + 5y - 4x = 5$ which simplifies to $-3x + 5y = 5$.Solving these two equations together can be done through various methods, but a quick observation or substitution will show that $x = 0$, $y = 1$, and substituting $x = 0$ back into $z = -4x$ gives $z = 0$. Therefore, the solution is $(x, y, z) = (0, 1, 0)$.

Solve the System {x−y+3z=−14x+z=0x+5y+z=5\left\{ \begin{array} { r r } x - y + 3 z & = - 1 \\4 x + z & = 0 \\x + 5 y + z & = 5\end{array} \right.⎩⎨⎧​x−y+3z4x+zx+5y+z​=−1=0=5​

Solve the System $\left\{ \begin{array} { r r } x - y + 3 z & = - 1 \\4 x + z & = 0 \\x + 5 y + z & = 5\end{array} \right.$