Solve the system by the substitution method.
-$$\left\{ \begin{array} { r r }
9 x + y = & 0 \
- 9 x + y = & - 18
\end{array} \right.$$
A) $\varnothing$
B) $\{ ( 1 , - 9 ) \}$
C) $\{ ( x , y ) \mid 9 x + y = 0 \}$
D) $\{ ( - 1,9 ) \}$

Question

Solve the system by the substitution method.
-$$\left\{ \begin{array} { r r } 
9 x + y = & 0 \
- 9 x + y = & - 18
\end{array} \right.$$
A) $\varnothing$
B) $\{ ( 1 , - 9 ) \}$
C) $\{ ( x , y ) \mid 9 x + y = 0 \}$
D) $\{ ( - 1,9 ) \}$

Quizplus · Accepted Answer

Solving the first equation for $y$, we get $y = -9x$. Substituting $y = -9x$ into the second equation gives $-9x + (-9x) = -18$, which simplifies to $-18x = -18$. Solving for $x$ gives $x = 1$. Substituting $x = 1$ back into $y = -9x$ gives $y = -9$. Therefore, the solution is $(1, -9)$.

Solve the System by the Substitution Method {9x+y=0−9x+y=−18\left\{ \begin{array} { r r } 9 x + y = & 0 \\- 9 x + y = & - 18\end{array} \right.{9x+y=−9x+y=​0−18​

Solve the System by the Substitution Method $\left\{ \begin{array} { r r } 9 x + y = & 0 \\- 9 x + y = & - 18\end{array} \right.$