After substituting, you should get two inequalities that each involve only $a$. Solve each one step by step (add or subtract first, then divide if needed).

The correct $a$ must satisfy both inequalities at the same time. After you find the range of $a$ that works, check which answer choices fall inside that range.

In Desmos, enter the three equations:

• y = 4x - 10
• y = x + 7
• y = 5

These are the two boundary lines from the inequalities and the horizontal line that corresponds to $y = 5$ for the point $(a,5)$.

Use Desmos’s intersection tool (tap where the lines meet) to find the intersection of y = 5 with y = 4x - 10, and also with y = x + 7. Note the x-coordinates of these two intersection points; those x-values are the endpoints of the interval of $a$-values that make $y = 5$ equal to each boundary.

On the graph, look along the horizontal line y = 5 between those two intersection x-values. In this region, y = 5 is above the line y = 4x - 10 and below the line y = x + 7, matching the inequalities $y > 4x - 10$ and $y < x + 7$. Check which answer choices have x-values lying between those two intersection x-values.

The point $(a,5)$ means $x=a$ and $y=5$.

Substitute these into both inequalities:

• From $y > 4x - 10$ you get

• From $y < x + 7$ you get

Now you have two inequalities involving only $a$.

Solve $5 > 4a - 10$ step by step:

Divide both sides by $4$:

This is the same as $a < \frac{15}{4}$ (about $3.75$).

Now solve $5 < a + 7$:

This tells you $a$ must be greater than $-2$.

Both inequalities must be true at the same time, so $a$ has to satisfy both:

• $a < \frac{15}{4}$
• $a > -2$

Together, this gives the interval

Now check each answer choice:

• -3 is not greater than $-2$.
• -1 is between $-2$ and $\frac{15}{4}$.
• 4 and 6 are not less than $\frac{15}{4}$.

So the only possible value of $a$ from the choices is $-1$.