Can we find an integer solution to three simultaneous inequalities?

Can we find an integer solution to three simultaneous inequalities?

Question

From the inequalities $y-2x>0,\qquad x+y>3,\qquad 2y-x<5$ deduce that $\frac{1}{3}< x < \frac{5}{3},\qquad 2< y <\frac{10}{3},$ and hence that the given inequalities cannot be satisfied simultaneously by integral values of $x$ and $y$.

[The phrase ‘integral values’ means the same as ‘integer values’. ]

Suggestion

Where do the three lines

$$\begin{align*} y-2x &= 0 \\ x+y &=3 \\ 2y-x &=5 \end{align*}$$

lie in the plane?

On which side of each of these lines does the respective inequality hold?

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

Which integer values of $x$ and $y$ do we need to consider?

Solution

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.

We can see from our sketch that there is a small triangular region satisfied by all three inequalities. We determine where each pair of lines intersects in order to determine the coordinates of the vertices of this triangle.

$$\begin{align*} y=2x\quad&\text{and}\quad x+y=3 \quad &\Longrightarrow \quad&& x&=1,& y&=2 \\ y=2x\quad&\text{and}\quad 2y-x=5 \quad &\Longrightarrow \quad&& x&=\tfrac53,& y&=\tfrac{10}{3} \\ x+y=3\quad&\text{and}\quad 2y-x=5 \quad &\Longrightarrow \quad&& x&=\tfrac13,& y&=\tfrac{8}{3} \end{align*}$$

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.