Solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. $$\left( \begin{array} { c }
y = 2 x + 32 \
y = - 6 x - 48
\end{array} \right)$$
A) ( -12, 10)
B) ( -10, 12)
C) ( 10, 12)
D) infinitely many solutions
E) no solutions

Question

Solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. $$\left( \begin{array} { c } 
y = 2 x + 32 \
y = - 6 x - 48
\end{array} \right)$$
A) ( -12, 10)
B) ( -10, 12)
C) ( 10, 12)
D) infinitely many solutions
E) no solutions

Quizplus · Accepted Answer

To solve the system, set the two equations equal to each other: $2x + 32 = -6x - 48$. Solving for $x$, we get $x = -10$. Substitute $x = -10$ into either equation to find $y$. Using $y = 2x + 32$, we get $y = 12$. Thus, the solution is $(-10, 12)$.

Solve Each System by Using Either the Substitution or the Elimination-By-Addition