Solve the system by the substitution method or the addition method.
-$\left\{ \begin{array} { l }
3 x + y = 5 \
4 x + 3 y = 0
\end{array} \right.$
A) $( 4,3 )$
B) $( 9 , - 5 )$
C) $( 12,3 )$
D) $( 3 , - 4 )$

Question

Solve the system by the substitution method or the addition method.
-$\left\{ \begin{array} { l } 
3 x + y = 5 \
4 x + 3 y = 0
\end{array} \right.$
A) $( 4,3 )$
B) $( 9 , - 5 )$
C) $( 12,3 )$
D) $( 3 , - 4 )$

Quizplus · Accepted Answer

Solving the system by substitution or addition, we first solve the first equation for y: $y = 5 - 3x$. Substituting this into the second equation gives $4x + 3(5 - 3x) = 0$, which simplifies to $4x + 15 - 9x = 0$, leading to $-5x = -15$, so $x = 3$. Substituting $x = 3$ back into $y = 5 - 3x$ gives $y = 5 - 3(3) = -4$. Therefore, the solution is $(3, -4)$.

Solve the System by the Substitution Method or the Addition {3x+y=54x+3y=0\left\{ \begin{array} { l } 3 x + y = 5 \\4 x + 3 y = 0\end{array} \right.{3x+y=54x+3y=0​

Solve the System by the Substitution Method or the Addition $\left\{ \begin{array} { l } 3 x + y = 5 \\4 x + 3 y = 0\end{array} \right.$