Solve the system of equations.
-$$\begin{array} { r }
x _ { 1 } + x _ { 2 } + x _ { 3 } = 6 \
x _ { 1 } = x _ { 3 } = - 2 \
x _ { 2 } + 3 x _ { 3 } = 11
\end{array}$$
A) $( 1,2,3 )$
B) No solution
C) $( - 1,2 , - 3 )$
D) $( 0,1,2 )$

Question

Solve the system of equations.
-$$\begin{array} { r } 
x _ { 1 } + x _ { 2 } + x _ { 3 } = 6 \
x _ { 1 } = x _ { 3 } = - 2 \
x _ { 2 } + 3 x _ { 3 } = 11
\end{array}$$
A) $( 1,2,3 )$
B) No solution
C) $( - 1,2 , - 3 )$
D) $( 0,1,2 )$

Quizplus · Accepted Answer

The first equation gives $x_1 + x_2 + x_3 = 6$. The second equation gives $x_1 = -2$ and $x_3 = -2$. Substituting these values into the first equation gives $x_2 = 6$. Therefore, the solution to the system of equations is $(x_1, x_2, x_3) = (-2, 6, -2)$.

Solve the System of Equations x1+x2+x3=6x1=x3=−2x2+3x3=11\begin{array} { r } x _ { 1 } + x _ { 2 } + x _ { 3 } = 6 \\x _ { 1 } = x _ { 3 } = - 2 \\x _ { 2 } + 3 x _ { 3 } = 11\end{array}x1​+x2​+x3​=6x1​=x3​=−2x2​+3x3​=11​

Solve the System of Equations $\begin{array} { r } x _ { 1 } + x _ { 2 } + x _ { 3 } = 6 \\x _ { 1 } = x _ { 3 } = - 2 \\x _ { 2 } + 3 x _ { 3 } = 11\end{array}$