If $a$ must be greater than one expression and less than another, what relationship must those two expressions have to each other for such an $a$ to exist?

After you write an inequality in $b$ from the condition in Hint 2, solve it carefully. Then remember: $b$ must be an integer, and the inequalities are strict, so equality at the boundary does not produce a solution.

In Desmos, let $x$ represent $a$ and $y$ represent $b$. Enter the two inequalities:

• 2x - y < 7
• x + 5y > 40 Desmos will shade the regions that satisfy each inequality; the overlap is where the system has solutions.

To check a specific integer value of $b$, graph a horizontal line such as y = 5, y = 6, y = 7, etc. For each line, look to see whether it passes through the overlapping shaded region. If it does, that means there is at least one $x$ (one value of $a$) that works for that $b$.

Start with a value of $y$ that clearly does not intersect the overlapping region, then increase $y$ by 1 each time (testing y = 5, y = 6, y = 7, and so on). The smallest integer $y$-value for which the horizontal line intersects the overlapping shaded region is the least integer value of $b$ that makes the system solvable.

Start by solving each inequality for $a$ in terms of $b$.

From $2a - b < 7$:

• Add $b$ to both sides: $2a < 7 + b$
• Divide by $2$: $a < \dfrac{7 + b}{2}$

From $a + 5b > 40$:

• Subtract $5b$ from both sides: $a > 40 - 5b$

So $a$ must satisfy both

For there to be at least one real value of $a$ that works, there must be some number that is greater than $40 - 5b$ and less than $\dfrac{7 + b}{2}$.

That is only possible if the lower bound is strictly less than the upper bound:

(This uses a strict $<$ because $a$ must be strictly greater than $40 - 5b$ and strictly less than $\dfrac{7 + b}{2}$.)

Now solve

Multiply both sides by $2$:

Move terms to group $b$ on one side and constants on the other:

So we get

The inequality $b > \dfrac{73}{11}$ means $b$ must be greater than $\dfrac{73}{11}$, not equal.

Compute or estimate $\dfrac{73}{11}$:

• $11 \times 6 = 66$
• $11 \times 7 = 77$

So $\dfrac{73}{11}$ is between $6$ and $7$ (about $6.64$). The smallest integer greater than this is $7$.

Therefore, the least integer value that $b$ can take so that the system has a solution is $7$.