Solve the system of equations by the elimination method.
-$$\left\{ \begin{array} { l }
4 x + y = 19 \
3 x + 4 y = - 2
\end{array} \right.$$
A) $( - 5,6 )$
B) $( 6 , - 5 )$
C) $( 0 , - 5 )$
D) $\varnothing$

Question

Solve the system of equations by the elimination method.
-$$\left\{ \begin{array} { l } 
4 x + y = 19 \
3 x + 4 y = - 2
\end{array} \right.$$
A) $( - 5,6 )$
B) $( 6 , - 5 )$
C) $( 0 , - 5 )$
D) $\varnothing$

Quizplus · Accepted Answer

To solve the system by elimination, we can multiply the first equation by 4 to make the coefficients of $y$ in both equations equal. This gives us $16x + 4y = 76$. Subtracting the second equation from this result eliminates $y$, leading to $13x = 78$, so $x = 6$. Substituting $x = 6$ into the first equation gives $4(6) + y = 19$, so $y = -5$. Thus, the solution is $(6, -5)$.

Solve the System of Equations by the Elimination Method {4x+y=193x+4y=−2\left\{ \begin{array} { l } 4 x + y = 19 \\3 x + 4 y = - 2\end{array} \right.{4x+y=193x+4y=−2​

Solve the System of Equations by the Elimination Method $\left\{ \begin{array} { l } 4 x + y = 19 \\3 x + 4 y = - 2\end{array} \right.$