Solve the system.
-$$\left\{ \begin{array} { c }
x + 5 y = 4 \
3 x + 15 y = 12
\end{array} \right.$$
A) $( 0,0 )$
B) $( - 3,0 )$
C) inconsistent (no solution)
D) $y = - \frac { x } { 5 } + \frac { 4 } { 5 }$, where $x$ is any real number

Question

Solve the system.
-$$\left\{ \begin{array} { c } 
x + 5 y = 4 \
3 x + 15 y = 12
\end{array} \right.$$
A) $( 0,0 )$
B) $( - 3,0 )$
C) inconsistent (no solution)
D) $y = - \frac { x } { 5 } + \frac { 4 } { 5 }$, where $x$ is any real number

Quizplus · Accepted Answer

The second equation is simply 3 times the first equation, which means they represent the same line. Therefore, there are infinitely many solutions, and any point on the line $y = - \frac { x } { 5 } + \frac { 4 } { 5 }$ is a solution.

Solve the System {x+5y=43x+15y=12\left\{ \begin{array} { c } x + 5 y = 4 \\3 x + 15 y = 12\end{array} \right.{x+5y=43x+15y=12​

Solve the System $\left\{ \begin{array} { c } x + 5 y = 4 \\3 x + 15 y = 12\end{array} \right.$