Solve the system by the addition method.
-$$\left\{ \begin{array} { r }
3 x + y = 10 \
12 x + 4 y = 40
\end{array} \right.$$
A) $\{ ( 0,10 ) \}$
B) $\varnothing$
C) $\{ ( 5 , - 5 ) \}$
D) $\{ ( x , y ) \mid 3 x + y = 10 \}$

Question

Solve the system by the addition method.
-$$\left\{ \begin{array} { r } 
3 x + y = 10 \
12 x + 4 y = 40
\end{array} \right.$$
A) $\{ ( 0,10 ) \}$
B) $\varnothing$
C) $\{ ( 5 , - 5 ) \}$
D) $\{ ( x , y ) \mid 3 x + y = 10 \}$

Quizplus · Accepted Answer

The second equation is a multiple of the first (multiply the first equation by 4 to get the second). This means the system has infinitely many solutions, all lying on the line described by the first equation, $3x + y = 10$.

Solve the System by the Addition Method {3x+y=1012x+4y=40\left\{ \begin{array} { r } 3 x + y = 10 \\12 x + 4 y = 40\end{array} \right.{3x+y=1012x+4y=40​

Solve the System by the Addition Method $\left\{ \begin{array} { r } 3 x + y = 10 \\12 x + 4 y = 40\end{array} \right.$