Linear Equation in 2 variables

$Let\quad \frac { x }{ y } \quad be\quad the\quad fraction\quad \\ 2y+1=x\quad \quad \Rightarrow \quad x-2y=1\quad ...(i)\\ \frac { x+2 }{ y+3 } =\frac { 3 }{ 2 } \Rightarrow 2x-3y=5\quad ...(ii)\\ \\ Multiplying\quad (i)\quad by\quad 2\quad (elimination\quad method)\\ and\quad then\quad subtracting\quad it\quad by\quad (ii)\\ 2x\quad -4y\quad =\quad 2\\ 2x\quad -3y\quad =\quad 5\\ \ominus \quad \oplus \quad \quad \quad \quad \ominus \quad \quad \quad \quad (changing\quad of\quad signs)\\ 0\quad -y\quad \quad \quad =-3\\ \\ y=3\\ \\ Substituting\quad the\quad value\quad of\quad y\quad in\quad (ii)\\ 2x-9=5\\ 2x=14\\ x=7\\ \\ Hence\quad the\quad fraction\quad is\quad \frac { 7 }{ 3 } \\ And\quad the\quad product\quad of\quad x\quad and\quad y\quad is\quad 21$

$x=2y+1 \Rightarrow \frac{x+2}{y+3}=\frac{2y+1+2}{y+3}=\frac{2}{3}$ Cross multiplying gives: $3(y+3)=2(2y+3) \Rightarrow3y+9=4y+6 \Rightarrow y=3$ Substitute y into $x=2y+1$ to give $x=7$ $\rightarrow 7 \times 3 = \boxed{21}$

This can be solved as a single variable equation!!