Use the elimination-by-addition method to solve each system. Match each system with the corresponding solution.
-$( 0 , - 4 )$
A) $\left( \begin{array} { c }
4 x - 6 y = 14 \
4 x + 2 y = - 10
\end{array} \right)$
B) $\left( \begin{array} { c }
6 x + 3 y = - 12 \
6 x - 4 y = 16
\end{array} \right)$

Question

Use the elimination-by-addition method to solve each system. Match each system with the corresponding solution.
-$( 0 , - 4 )$ 
A) $\left( \begin{array} { c } 
4 x - 6 y = 14 \
4 x + 2 y = - 10
\end{array} \right)$ 
B) $\left( \begin{array} { c } 
6 x + 3 y = - 12 \
6 x - 4 y = 16
\end{array} \right)$

Quizplus · Accepted Answer

To solve the system using the elimination-by-addition method, we need to add or subtract the two equations in a way that eliminates one of the variables. In system A, adding the two equations gives us $8x = 4$, which simplifies to $x = \frac{1}{2}$. Substituting this into either equation gives us $-6y = 14 - 2 = 12$, which simplifies to $y = -2$. Therefore, the solution to system A is $(\frac{1}{2}, -2)$, which does not match the given point.

In system B, adding the two equations gives us $12x = 4$, which simplifies to $x = \frac{1}{3}$. Substituting this into either equation gives us $-y = -\frac{4}{3}$, which simplifies to $y = \frac{4}{3}$. Therefore, the solution to system B is $(\frac{1}{3}, \frac{4}{3})$, which matches the given point $(0,-4)$. Thus, the best choice is system B.

Use the Elimination-By-Addition Method to Solve Each System (0,−4)( 0 , - 4 )(0,−4)

Use the Elimination-By-Addition Method to Solve Each System $( 0 , - 4 )$