Solve the system by using the substitution method. $$\left( \begin{array} { c }
x + 4 y = 3 \
3 x + 12 y = - 1
\end{array} \right)$$
A) $( 3,0 )$
B) $\{ ( x , y ) \mid x + 4 y = 3 \}$
C) $( 3 , - 4 )$
D) $( - 7,7 )$
E) inconsistent

Question

Solve the system by using the substitution method. $$\left( \begin{array} { c } 
x + 4 y = 3 \
3 x + 12 y = - 1
\end{array} \right)$$
A) $( 3,0 )$
B) $\{ ( x , y ) \mid x + 4 y = 3 \}$
C) $( 3 , - 4 )$
D) $( - 7,7 )$
E) inconsistent

Quizplus · Accepted Answer

The second equation is a multiple of the first (multiplied by 3), but the constants do not align (3*3 ≠ -1), indicating the system is inconsistent.

Solve the System by Using the Substitution Method (x+4y=33x+12y=−1)\left( \begin{array} { c } x + 4 y = 3 \\3 x + 12 y = - 1\end{array} \right)(x+4y=33x+12y=−1​)

Solve the System by Using the Substitution Method $\left( \begin{array} { c } x + 4 y = 3 \\3 x + 12 y = - 1\end{array} \right)$