Solve the system of equations by using the substitution method.
-$$\begin{array} { l }
3 ( x + y ) = - 7 - y \
9 x - 1 = 8 + 3 y
\end{array}$$
A) $\left\{ \left( - \frac { 7 } { 3 } , 0 \right) \right\}$
B) $\left\{ \left( - \frac { 7 } { 3 } , - \frac { 1 } { 3 } \right) \right\}$
C) $\left\{ \left( \frac { 1 } { 3 } , - 2 \right) \right\}$
D) $\left\{ \left( 0 , - \frac { 1 } { 3 } \right) \right\}$

Question

Solve the system of equations by using the substitution method.
-$$\begin{array} { l } 
3 ( x + y ) = - 7 - y \
9 x - 1 = 8 + 3 y
\end{array}$$
A) $\left\{ \left( - \frac { 7 } { 3 } , 0 \right) \right\}$
B) $\left\{ \left( - \frac { 7 } { 3 } , - \frac { 1 } { 3 } \right) \right\}$
C) $\left\{ \left( \frac { 1 } { 3 } , - 2 \right) \right\}$
D) $\left\{ \left( 0 , - \frac { 1 } { 3 } \right) \right\}$

Quizplus · Accepted Answer

First, solve the first equation for y: $3(x + y) = -7 - y$ simplifies to $3x + 4y = -7$. Solve for y: $y = -\frac{7}{4} - \frac{3}{4}x$. Substitute this into the second equation: $9x - 1 = 8 + 3(-\frac{7}{4} - \frac{3}{4}x)$. Simplify and solve for x: $x = \frac{1}{3}$. Substitute back to find y: $y = -2$. Thus, the solution is $\left( \frac{1}{3}, -2 \right)$.

Solve the System of Equations by Using the Substitution Method 3(x+y)=−7−y9x−1=8+3y\begin{array} { l } 3 ( x + y ) = - 7 - y \\9 x - 1 = 8 + 3 y\end{array}3(x+y)=−7−y9x−1=8+3y​

Solve the System of Equations by Using the Substitution Method $\begin{array} { l } 3 ( x + y ) = - 7 - y \\9 x - 1 = 8 + 3 y\end{array}$