Use the elimination method to solve the system.
-$\left\{ \begin{array} { c } 3 x + 48 y = 48 \ 3 x - 6 y = - 6 \ \end{array} \right.$

A) x = 0 , y = 1
B) $x = 1 , y = 1$
C) $x = 0 , y = 0$
D) $x = 1 , y = 0$

Question

Use the elimination method to solve the system.
-$\left\{ \begin{array} { c } 3 x + 48 y = 48 \ 3 x - 6 y = - 6 \ \end{array} \right.$

A) x = 0 , y = 1 
B) $x = 1 , y = 1$
C) $x = 0 , y = 0$
D) $x = 1 , y = 0$

Quizplus · Accepted Answer

Subtracting the second equation from the first eliminates $x$, leading to $54y = 54$, so $y = 1$. Substituting $y = 1$ into the second equation gives $3x - 6(1) = -6$, so $3x = 0$, and $x = 0$.

Use the Elimination Method to Solve the System {3x+48y=483x−6y=−6\left\{ \begin{array} { c } 3 x + 48 y = 48 \\ 3 x - 6 y = - 6 \\ \end{array} \right.{3x+48y=483x−6y=−6​

Use the Elimination Method to Solve the System $\left\{ \begin{array} { c } 3 x + 48 y = 48 \\ 3 x - 6 y = - 6 \\ \end{array} \right.$