From the inequalities \[y-2x>0,\qquad x+y>3,\qquad 2y-x<5\] deduce that \[\frac{1}{3}< x < \frac{5}{3},\qquad 2< y <\frac{10}{3}\]

Consider the three inequalities in the plane:

\(y-2x = 0\) crosses the axes at \((0,0)\) and has gradient \(2\) (it rearranges to \(y=2x\)). The region \(y-2x>0\) lies above this line, as it includes the point \((0,1)\).

\(x+y =3\) crosses the axes at \((0,3)\) and \((3,0)\). The region \(x+y>3\) lies above this line, as it does not include the point \((0,0)\).

\(2y-x =5\) crosses the axes at \((0,\frac52)\) and \((-5,0)\). The region \(2y-x<5\) lies below this line, as it includes the point \((0,0)\).

We can therefore sketch these three regions in the plane. Here the shaded regions are those which **do not** satisfy the inequality.

From these we can deduce the minumum and maximum extent of both \(x\) and \(y\): \[\frac{1}{3} < x < \frac{5}{3}, \qquad 2< y < \frac{10}{3},\] as required.

… and hence that the given inequalities cannot be satisfied simultaneously by integral values of \(x\) and \(y\).

The only integer value of \(x\) that satisfies \(\frac{1}{3} < x < \frac{5}{3}\) is \(x=1\).

The only integer value of \(y\) that satisfies \(2< y < \frac{10}{3}\) is \(y=3\).

Does the point \((1,3)\) obey all three inequalities simultaneously? \[\begin{align*} y-2x &> 0: \quad 3-2=1>0 \quad \checkmark \\ x+y &>3: \quad 1+3=4>3 \quad \checkmark \\ 2y-x &<5: \quad 6-1=5 \nless 5 \quad \mathsf{X} \end{align*}\]Thus there is no integer pair of \(x\) and \(y\) that satisfies all three inequalities simultaneously.