1. State the problem: Solve the system $2x+3y=7$ and $x-y=1$, then find $x+y$. 2. Write the system clearly: $$ \begin{cases} 2x+3y=7\\ x-y=1 \end{cases} $$ 3. Use the second equation to substitute for $x$. From $x-y=1$, add $y$ to both sides: $$x=1+y$$ 4. Substitute $x=1+y$ into the first equation $2x+3y=7$: $$2(1+y)+3y=7$$ 5. Simplify step by step: $$2(1+y)+3y=2+2y+3y=2+5y$$ So: $$2+5y=7$$ 6. Subtract $2$ from both sides (show the cancel line): $$2+5y-\cancel{2}=7-\cancel{2}$$ $$5y=5$$ 7. Divide both sides by $5$ (show the cancel line): $$\frac{\cancel{5}y}{\cancel{5}}=\frac{\cancel{5}}{\cancel{5}}$$ $$y=1$$ 8. Plug $y=1$ back into $x=1+y$: $$x=1+1=2$$ 9. Compute the requested value $x+y$: $$x+y=2+1=3$$ 10. Final answer: $\boxed{3}$.