Solve the system by using the elimination-by-addition method. $\left( \begin{array} { l }
3 x + 6 y = 3 \
5 x - 6 y = 1
\end{array} \right)$
A) $\left( - \frac { 1 } { 4 } , - \frac { 1 } { 4 } \right)$
B) $\left( \frac { 1 } { 4 } , \frac { 1 } { 4 } \right)$
C) $\left( - \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
D) $\left( \frac { 1 } { 2 } , - \frac { 1 } { 4 } \right)$
E) $\left( \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$

Question

Solve the system by using the elimination-by-addition method. $\left( \begin{array} { l } 
3 x + 6 y = 3 \
5 x - 6 y = 1
\end{array} \right)$
A) $\left( - \frac { 1 } { 4 } , - \frac { 1 } { 4 } \right)$
B) $\left( \frac { 1 } { 4 } , \frac { 1 } { 4 } \right)$
C) $\left( - \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
D) $\left( \frac { 1 } { 2 } , - \frac { 1 } { 4 } \right)$
E) $\left( \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$

Quizplus · Accepted Answer

Adding the two equations $3x + 6y = 3$ and $5x - 6y = 1$ eliminates $y$, resulting in $8x = 4$, which simplifies to $x = \frac{1}{2}$. Substituting $x = \frac{1}{2}$ into the first equation gives $3(\frac{1}{2}) + 6y = 3$, leading to $6y = 1.5$, so $y = \frac{1}{4}$. Therefore, the solution is $\left( \frac{1}{2}, \frac{1}{4} \right)$.

Solve the System by Using the Elimination-By-Addition Method (3x+6y=35x−6y=1)\left( \begin{array} { l } 3 x + 6 y = 3 \\5 x - 6 y = 1\end{array} \right)(3x+6y=35x−6y=1​)

Solve the System by Using the Elimination-By-Addition Method $\left( \begin{array} { l } 3 x + 6 y = 3 \\5 x - 6 y = 1\end{array} \right)$