[tex] \text{Given : 2x - y = 4 and x + 2y = 7}[/tex]
[tex] \text{Solution : according to the question here we are provided with} \\ \text{two equations separately. So, let's name them as eq(1) and eq(2).} \\ \text{ The rest is followed as we continue to get to our goal} \\ \\ \sf {1}^{st} \: equation : \\ \quad\dashrightarrow \frak{2x - y = 4} \\ \\ \quad \dashrightarrow \frak{ \red{x = \frac{y + 4}{2} }} \qquad \sf ...eq(1) \\ \rm{ now \: \: as \:per \: our \: plan \: we \: need \: to \: get \: eq(2) } \\ \\ \sf{ {2}^{nd} \:equation} : \\ \quad\dashrightarrow \frak{ x + 2y = 7} \\ \\ \quad\dashrightarrow \frak{ \red{y = \frac{7 - x}{2}}} \qquad \sf ...eq(2) \\ \\ \text{Now, finding x and y} \\ \\ \underline{ \frak{Finding \: x}}\\ \\ \quad \hookrightarrow \frak{x = \frac{y + 4}{2}} \\ \\ \quad \hookrightarrow \frak{2x = \frac{7 - x}{2} + 4} \\ \\ \quad \hookrightarrow \frak{2x = \frac{7 - x + 8}{2}} \\ \\ \quad \hookrightarrow \frak{2x = \frac{1 - x}{2}} \\ \\\quad \hookrightarrow \frak{4x = 1 - x} \\ \\ \quad \hookrightarrow \frak{5x = 1} \\ \\\quad \star \qquad \underline{ \boxed{ \green{ \frak{x = \frac{1}{5}}}}} \\ \\ \underline{ \frak{Finding \: y}} \\ \\ \quad \hookrightarrow \frak{y = \frac{7 - x}{2}} \\ \\ \hookrightarrow \frak{2y = 7 - \frac{1}{5} } \\ \\ \hookrightarrow \frak{2y = \frac{35 - 1}{5} } \\ \\ \hookrightarrow \frak{2y = \frac{34}{5}} \\ \\ \hookrightarrow \frak{y = \frac{34}{5} \times \frac{1}{2}} \\ \\ \hookrightarrow \frak{y = \frac{17}{5} } \\ \\ \star \qquad \underline{ \boxed{ \frak{ \green{y = 3 \frac{2}{5} }}}}[/tex]
