Question 119·Hard·Linear Inequalities in One or Two Variables

Try writing $x-y$ as a combination of $2x+y$ and $x+y$. Be careful: multiplying an inequality by a negative number reverses its direction.

After finding an upper bound for $x-y$, test whether a point can actually achieve that bound while also satisfying $x+2y\leq 10$.

Algebra is faster here because a combination of the inequalities gives a direct upper bound for $x-y$. Desmos can quickly verify the boundary point.

Enter the first three inequalities: $2x+y\leq 11$, $x+2y\leq 10$, and $x+y\geq 6$. Their overlapping shaded region is the set of points that satisfy those conditions.

Enter $x-y=k$ and use the slider for $k$. Increase $k$ until the line no longer touches the overlapping shaded region. The last point of contact is $(5,1)$, where the boundary line has its greatest possible value of $k$.

Rewrite the target expression using two of the given expressions:

Since $2x+y\leq 11$, multiplying by $2$ gives $2(2x+y)\leq 22$.

Since $x+y\geq 6$, multiplying by $-3$ reverses the inequality and gives $-3(x+y)\leq -18$.

Adding these results gives

Therefore, if $x-y\geq k$, then $k$ cannot be greater than $4$.

Test $(x,y)=(5,1)$ in the first three inequalities:

These values satisfy all three inequalities, and $x-y=5-1=4$. Thus, a solution exists when $k=4$, as well as for every smaller value of $k$.