Use the elimination-by-addition method to solve each system. Match each system with the corresponding solution.
-$( - 1 , - 3 )$
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.
-$( - 1 , - 3 )$ 
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 use the elimination-by-addition method, we need to manipulate the equations so that we can add them together and eliminate one of the variables. For system A, we can multiply the second equation by -3 and add it to the first equation to eliminate y:

\begin{align*}
4x - 6y &= 14 \
-12x - 6y &= 30 \
\hline
-8x &= 44
\end{align*}

Solving for x, we get x = -\frac{11}{2}. We can then substitute this value into one of the original equations (let's use the first one) to solve for y:

\begin{align*}
4\left(-\frac{11}{2}ight) - 6y &= 14 \
-22 - 6y &= 14 \
-6y &= 36 \
y &= -6
\end{align*}

Therefore, the solution to system A is (-1, -3). For system B, we can try the same method, but we won't be able to eliminate a variable using addition because the coefficients of x and y are always different.

Use the Elimination-By-Addition Method to Solve Each System (−1,−3)( - 1 , - 3 )(−1,−3)

Use the Elimination-By-Addition Method to Solve Each System $( - 1 , - 3 )$