Solve the system of equations by the substitution method.
-$\left\{ \begin{array} { l }
4 y = x + 32 \
3 x + 12 y = 0
\end{array} \right.$
A) $( 4 , - 16 )$
B) $( - 16,4 )$
C) no solution
D) infinite number of solutions

Question

Solve the system of equations by the substitution method.
-$\left\{ \begin{array} { l } 
4 y = x + 32 \
3 x + 12 y = 0
\end{array} \right.$
A) $( 4 , - 16 )$
B) $( - 16,4 )$
C) no solution
D) infinite number of solutions

Quizplus · Accepted Answer

From the first equation, we can express $x$ in terms of $y$ as $x = 4y - 32$. Substituting this expression for $x$ in the second equation gives $3(4y - 32) + 12y = 0$. Simplifying this equation leads to $12y - 96 + 12y = 0$, which further simplifies to $24y = 96$, giving $y = 4$. Substituting $y = 4$ back into the expression for $x$ yields $x = 4(4) - 32 = -16$. Therefore, the solution is $(-16, 4)$.

Solve the System of Equations by the Substitution Method {4y=x+323x+12y=0\left\{ \begin{array} { l } 4 y = x + 32 \\3 x + 12 y = 0\end{array} \right.{4y=x+323x+12y=0​

Solve the System of Equations by the Substitution Method $\left\{ \begin{array} { l } 4 y = x + 32 \\3 x + 12 y = 0\end{array} \right.$