Solve Linear Systems by Substitution
-$$\begin{array} { l }
3 x - 2 y = - 28 - 3 x \
4 x + 7 y = x + 6 y + - 4
\end{array}$$
A) $\{ ( - 3,5 ) \}$
B) $\{ ( - 3 , - 5 ) \}$
C) $\{ ( 0 , - 28 ) \}$
D) $\left\{ \left( - \frac { 14 } { 3 } , 0 \right) \right\}$

Question

Solve Linear Systems by Substitution
-$$\begin{array} { l } 
3 x - 2 y = - 28 - 3 x \
4 x + 7 y = x + 6 y + - 4
\end{array}$$
A) $\{ ( - 3,5 ) \}$
B) $\{ ( - 3 , - 5 ) \}$
C) $\{ ( 0 , - 28 ) \}$
D) $\left\{ \left( - \frac { 14 } { 3 } , 0 \right) \right\}$

Quizplus · Accepted Answer

First, simplify both equations: 1) $6x - 2y = -28$, 2) $3x + y = -4$. Solving the second equation for $y$ gives $y = -4 - 3x$. Substituting $y$ in the first equation and solving for $x$ gives $x = -3$. Substituting $x = -3$ back into $y = -4 - 3x$ gives $y = 5$. Therefore, the solution is $(-3, 5)$.

Solve Linear Systems by Substitution - 3x−2y=−28−3x4x+7y=x+6y+−4\begin{array} { l } 3 x - 2 y = - 28 - 3 x \\4 x + 7 y = x + 6 y + - 4\end{array}3x−2y=−28−3x4x+7y=x+6y+−4​

Solve Linear Systems by Substitution
- $\begin{array} { l } 3 x - 2 y = - 28 - 3 x \\4 x + 7 y = x + 6 y + - 4\end{array}$