Solve the problem.
-$$\left\{ \begin{array} { r }
x - 2 y = 16 \
- 2 x - 3 y = 3
\end{array} \right.$$
A) no solution; $\varnothing$
B) $\{ ( 6 , - 5 ) \}$
C) $\{ ( 5 , - 4 ) \}$
D) $\{ ( - 6 , - 4 ) \}$

Question

Solve the problem.
-$$\left\{ \begin{array} { r } 
x - 2 y = 16 \
- 2 x - 3 y = 3
\end{array} \right.$$
A) no solution; $\varnothing$
B) $\{ ( 6 , - 5 ) \}$
C) $\{ ( 5 , - 4 ) \}$
D) $\{ ( - 6 , - 4 ) \}$

Quizplus · Accepted Answer

To solve the system of equations, we can use the method of substitution or elimination. Here, using elimination might be more straightforward. We can multiply the first equation by 2 to make the coefficients of $x$ in both equations equal in magnitude but opposite in sign, which makes it easier to eliminate $x$ when we add the two equations together.First equation after multiplying by 2:$$2(x - 2y) = 2(16)$$$$2x - 4y = 32$$Now, adding this to the second equation:$$2x - 4y + (-2x - 3y) = 32 + 3$$$$2x - 2x - 4y - 3y = 35$$$$-7y = 35$$$$y = -5$$Substitute $y = -5$ into the first original equation to solve for $x$:$$x - 2(-5) = 16$$$$x + 10 = 16$$$$x = 6$$Therefore, the solution is $(x, y) = (6, -5)$, which corresponds to choice B.

Solve the Problem {x−2y=16−2x−3y=3\left\{ \begin{array} { r } x - 2 y = 16 \\- 2 x - 3 y = 3\end{array} \right.{x−2y=16−2x−3y=3​

Solve the Problem $\left\{ \begin{array} { r } x - 2 y = 16 \\- 2 x - 3 y = 3\end{array} \right.$