David and Gabe are working on a group paper. However, since both of them are also busy doing other things, one usually starts writing the paper ahead of the other. If David starts writing the paper by himself and Gabe joins in 2 hours later, they can finish the paper in 5 hours. If Gabe starts writing the paper by himself and David joins in 2 hours later, they can finish the paper in 6 hours. How long will take to finish the paper if they start at the same time?
The system of equation that I used here is $\frac2D+\frac3D+\frac3G=\frac15$ and $\frac2G+\frac4G+\frac4D=\frac16$. The answer I got is very far from the correct answer.
Say David writes $x$ (a proportion) of the paper in one hour and Gabe writes $y$ of it in one hour. Then the statements give $$2x+3(x+y)=1\implies 5x+3y=1$$ $$2y+4(x+y)=1\implies 4x+6y=1$$ Solving these equations gives $x=\frac16$ and $y=\frac1{18}$. The final answer is then the $k$ that satisfies $k(x+y)=1$, so $k=\frac1{x+y}=\frac1{1/6+1/18}=\frac92$ or $4.5$ hours.