Answer:
The ordered pair [tex](-2,\, -4)[/tex] is indeed a solution to the system:
[tex]\left\lbrace\begin{aligned} & y = \frac{1}{2}\, x - 3 \\ & y = -2\, x - 8\end{aligned}\right.[/tex].
Step-by-step explanation:
Consider a system of equations about variables [tex]x[/tex] and [tex]y[/tex]. An ordered pair [tex](x_{0},\, y_{0})[/tex] (where [tex]x_{0}[/tex] and [tex]y_{0}[/tex] are constant) is a solution to that system if and only if all equations in that system hold after substituting in [tex]x = x_{0}[/tex] and [tex]y = y_{0}[/tex].
For the system in this question, [tex](-2,\, -4)[/tex] would be a solution only if both equations in the system hold after replacing all [tex]x[/tex] in equations of the system with [tex](-2)[/tex] and all [tex]y[/tex] with [tex](-4)[/tex].
The [tex]\texttt{LHS}[/tex] of the equation [tex]y = (1/2)\, x - 3[/tex] would become [tex](-4)[/tex]. The [tex]\texttt{RHS}[/tex] of that equation would become [tex](1/2) \, (-2) - 3[/tex]. The two sides are indeed equal.
Similarly, the [tex]\texttt{LHS}[/tex] of the equation [tex]y = -2\, x - 8[/tex] would become [tex](-4)[/tex]. The [tex]\texttt{RHS}[/tex] of that equation would become [tex](-2)\, (-2) - 8[/tex]. The two sides are indeed equal.
Thus, [tex]x = (-2)[/tex] and [tex]y = (-4)[/tex] simultaneously satisfy both equations of the given system. Therefore, the ordered pair [tex](-2,\, -4)[/tex] would indeed be a solution to that system.
