Question 450·Hard·Algebra

Type $2x+3y\le15$ into Desmos. It shades below the solid boundary line $2x+3y=15$: for a fixed $x$, increasing $y$ makes the left side larger. The line is solid because $\le$ allows equality.

Add $y\ge x+4$. Desmos now shows the overlap above this line and below the first. Its rightmost edge is where the solid boundaries cross, between $x=0$ and $x=1$. Farther right, the required minimum $y$ rises while the allowed maximum $y$ falls, so no point can satisfy both conditions.

The greatest integer to the left of that edge is $0$. At $x=0$, the second boundary gives $y=0+4=4$. Type $(0,4)$; Desmos plots it in both shadings, so this value of $x$ is possible. The overlap stops before $x=1$, so no greater integer works. Use the edge to limit $x$, then check a point at your integer. The greatest integer value of $x$ is $0$. Grid in 0.

For any $x$, the smallest allowed $y$ is $x+4$. Because the first inequality contains positive $3y$, any larger $y$ makes it harder to satisfy. So some $y$ can work exactly when that smallest one works. Substitute $y=x+4$:

Multiply $3$ by each term inside the parentheses:

The $12$ comes from $3\cdot4$.

Combine $2x$ and $3x$ to get one $x$ term:

Subtract $12$ from both sides; subtraction keeps the inequality direction:

Divide both sides by positive $5$, so the inequality direction stays the same:

Since $\frac{3}{5}$ lies between $0$ and $1$, no positive integer $x$ can work. The point $(0,4)$ from the graph confirms that $x=0$ does work. So the greatest integer value of $x$ is $0$. Grid in 0.