[tex] \text{Given : Which ordered pair is the solution to the system of equations?} \\ \\ \begin{cases}-5x - 3y = 12 \\ y - x = 12 \end{cases} [/tex]
[tex] \text{Solution : According to the question, here let's take these variables as two} \\ \text{equations then get their value and finally get the answer.} \\ \\ \sf {1}^{st} \: equation : \\ \leadsto \frak{-5x - 3y = 12} \\ \\ \leadsto \frak{-5x = 12 + 3y} \\ \\ \leadsto \frak{x= \frac{12 + 3y}{-5}} \qquad \sf ...eq(1) \\ \\ \sf 2^{nd} \: equation : \\ \leadsto \frak{y = x - 12} \qquad \sf ...eq(2) \\ \\ \underline{ \frak{finding \: x }} : \\ \\ \dashrightarrow \frak{x= \frac{12 + 3y}{-5}} \\ \\ \dashrightarrow \frak{x= \frac{12 + 3(x - 12)}{-5}} \\ \\ \dashrightarrow \frak{x= \frac{12 + 3x - 36}{-5}} \\ \\ \dashrightarrow \frak{-5x= 12 + 3x - 36} \\ \\ \dashrightarrow \frak{-5x - 3x= - 24} \\ \\ \dashrightarrow \frak{-8x = - 24} \\ \\ \star \quad \underline{ \boxed{ \green{ \frak{x = 3}}}} \\ \\ \underline{ \frak{ getting \: y}} : \\ \\ \dashrightarrow \frak{y = 3 - 12} \\ \\ \star \quad\underline{ \boxed{\blue{ \frak{y = - 9}}}} \\ \\ [/tex]
[tex] \green{ \mathbb{HENCEFORTH, \: THE \: PAIR \: IS \: \pink{\frak{ (3,-9)}}}}[/tex]
