Solve Linear Systems by Substitution
-$$\begin{array} { l }
y = \frac { 1 } { 6 } x - 7 \
y = \frac { 2 } { 3 } x + 11
\end{array}$$
A) $\{ ( - 36 , - 13 ) \}$
B) $\{ ( - 36,13 ) \}$
C) $\{ ( 42,0 ) \}$
D) $\left\{ \left( - \frac { 35 } { 6 } , - \frac { 53 } { 3 } \right) \right\}$

Question

Solve Linear Systems by Substitution
-$$\begin{array} { l } 
y = \frac { 1 } { 6 } x - 7 \
y = \frac { 2 } { 3 } x + 11
\end{array}$$
A) $\{ ( - 36 , - 13 ) \}$
B) $\{ ( - 36,13 ) \}$
C) $\{ ( 42,0 ) \}$
D) $\left\{ \left( - \frac { 35 } { 6 } , - \frac { 53 } { 3 } \right) \right\}$

Quizplus · Accepted Answer

To solve the system by substitution, set the two expressions for $y$ equal to each other: $\frac{1}{6}x - 7 = \frac{2}{3}x + 11$. Solving for $x$ gives $x = -36$. Substituting $x = -36$ into either equation to solve for $y$ gives $y = -13$. Therefore, the solution is $(-36, -13)$.

Solve Linear Systems by Substitution - y=16x−7y=23x+11\begin{array} { l } y = \frac { 1 } { 6 } x - 7 \\y = \frac { 2 } { 3 } x + 11\end{array}y=61​x−7y=32​x+11​

Solve Linear Systems by Substitution
- $\begin{array} { l } y = \frac { 1 } { 6 } x - 7 \\y = \frac { 2 } { 3 } x + 11\end{array}$