Solving Inequalities for Positive Integers $p$ and $q$

We are given two inequalities with the condition that $p$ and $q$ are positive integers ($p \ge 1$, $q \ge 1$).

• Inequality 1: $p^2 - 4q < 4$
• Inequality 2: $3p + 2q < 6$

Analyze Inequality 2 for Integer Solutions

From the second inequality, $3p + 2q < 6$. Since $p$ and $q$ must be positive integers ($p \ge 1, q \ge 1$):

• If $p = 1$, then $3(1) + 2q < 6 \implies 3 + 2q < 6 \implies 2q < 3$. The only positive integer value for $q$ satisfying this is $q = 1$.
• If $p = 2$, then $3(2) + 2q < 6 \implies 6 + 2q < 6 \implies 2q < 0$. There are no positive integer values for $q$.
• If $p \ge 2$, $3p$ will be $6$ or greater, making $3p + 2q < 6$ impossible for positive $q$.

Therefore, the only possible integer pair satisfying $3p + 2q < 6$ with $p \ge 1$ and $q \ge 1$ is $p = 1$ and $q = 1$.

Verify Solution with Inequality 1

Now, we check if the pair $(p=1, q=1)$ satisfies the first inequality, $p^2 - 4q < 4$.

• Substitute $p=1$ and $q=1$: $(1)^2 - 4(1) < 4$
• Calculate: $1 - 4 < 4$
• Result: $-3 < 4$. This is true.

The pair $(p=1, q=1)$ satisfies both inequalities.

Calculate the Value of $p+q$

Using the determined values $p=1$ and $q=1$:

• $p + q = 1 + 1 = 2$

The value of $(p + q)$ is 2.