Consider the following inequalities

                                                                  𝑝² − 4𝑞 < 4

                                                                 3𝑝 + 2𝑞 < 6

where 𝑝 and 𝑞 are positive integers.

The value of (𝑝 + 𝑞) is _______.

The correct answer is 2.

As shown in the provided image, we are given two inequalities involving positive integers $p$ and $q$:
1) Inequality 1: $p^2-4q<4$
2) Inequality 2: $3p+2q<6$

Let's analyze and combine these inequalities step-by-step to find the positive integer values of $p$ and $q$.

Step 1: Rearrange Inequality 1
We can isolate the term containing $q$ on one side. By adding $4q$ and subtracting $4$ from both sides, we get:

$$p^2-4<4q$$

Let this be Inequality (i).

Step 2: Rearrange Inequality 2
To align the $q$ terms so they can be eliminated or compared easily with Inequality (i), we can multiply the second inequality by $2$:

$$2(3p+2q)<2\left(6\right)$$

$$6p+4q<12$$

Now, isolate the term $-4q$ by subtracting $4q$ and $12$ from both sides:

$$6p-12<-4q$$

Let this be Inequality (ii).

Step 3: Combine Inequality (i) and Inequality (ii)
Adding Inequality (i) and Inequality (ii) side-by-side allows us to eliminate the $q$ terms:

$$(p^2-4)+(6p-12)<4q+(-4q)$$

$$p^2+6p-16<0$$

Step 4: Factor and solve the quadratic inequality for $p$
We can factor the quadratic expression by finding two numbers that multiply to $-16$ and add to $6$. These numbers are $8$ and $-2$:

$$(p+8)(p-2)<0$$

This inequality holds true when the variable $p$ lies between the two roots of the quadratic equation:

$$p\in (-8,2)$$

Since the problem states that $p$ must be a positive integer, the only possible value for $p$ in the interval $(-8,2)$ is:

$$p=1$$

Step 5: Solve for $q$
Now we substitute $p=1$ back into our inequalities to determine the value of $q$.

From Inequality (i):

$$1^2-4<4q$$

$$-3<4q\Rightarrow q>-\frac{3}{4}$$

From Inequality (ii) or directly from the original Inequality 2:

$$3\left(1\right)+2q<6$$

$$3+2q<6\Rightarrow 2q<3\Rightarrow q<\frac{3}{2}$$

Combining these bounds for $q$ gives:

$$-\frac{3}{4}<q<\frac{3}{2}$$

Since $q$ is also a positive integer, the only integer in this interval is:

$$q=1$$

Step 6: Calculate $p+q$
Using our values $p=1$ and $q=1$:

$$p+q=1+1=2$$