Finding Integer Solutions To Equations A Math Adventure

Hey guys! Today, we're diving into a fun little math problem where we need to find a possible value for $p$, given that the solution $x$ to the equation $4x = 12 - px$ is an integer, and $p$ is a positive integer. Sounds like a puzzle, right? Let's break it down and solve it together!

Diving into the Equation: Unraveling the Mystery of $4x = 12 - px$

Our primary mission is to decipher the equation $4x = 12 - px$. This equation holds the key to our quest for finding a suitable value for $p$. To kick things off, let's gather all the $x$ terms on one side. This is a classic algebraic maneuver that helps us isolate the variable we're interested in. By adding $px$ to both sides of the equation, we achieve a pivotal transformation:

$$4x + px = 12$$

Now, we spot a golden opportunity to factor out $x$ from the left side. Factoring is like reverse distribution, and it's a powerful tool for simplifying expressions. By factoring out $x$, we reshape our equation into a more manageable form:

$$x(4 + p) = 12$$

This transformation is a game-changer! It brings us closer to understanding the relationship between $x$ and $p$. Now, to truly isolate $x$, we'll divide both sides of the equation by $(4 + p)$. This step will reveal $x$ in terms of $p$:

$$x = \frac{12}{4 + p}$$

The Integer Constraint: Why It Matters

This equation, $x = \frac12}{4 + p}$, is a crucial stepping stone in our problem-solving journey. It explicitly expresses $x$ in terms of $p$. But here's the twist we're not looking for just any value of $x$; we're seeking an integer solution. This integer constraint is a critical piece of the puzzle. It dictates that the result of the division, $\frac{12{4 + p}$, must be a whole number—no fractions or decimals allowed!

Why is this integer constraint so important? Because it significantly narrows down the possibilities for $p$. If $x$ must be an integer, then $(4 + p)$ must be a factor of 12. In other words, $(4 + p)$ must divide 12 evenly, leaving no remainder. This realization is a major breakthrough in our quest to find a suitable value for $p$. It transforms the problem from an open-ended search to a focused exploration of factors.

Factors of 12

So, what are the factors of 12? Well, the factors of 12 are the numbers that divide 12 without leaving a remainder. Let's list them out:

• 1
• 2
• 3
• 4
• 6
• 12

These numbers are our candidates for the value of $(4 + p)$. Each of them offers a potential pathway to an integer solution for $x$. But remember, our goal isn't just to find any factor; we need to find a factor that leads to a valid value for $p$.

Hunting for $p$: Decoding the Possible Values

Now comes the exciting part: the hunt for $p$. We know that $(4 + p)$ must be one of the factors of 12 that we identified earlier. So, let's explore each possibility and see what values of $p$ they yield. Remember, $p$ is a positive integer, so we need to keep that constraint in mind as we proceed.

We'll methodically analyze each factor of 12, setting $(4 + p)$ equal to each factor and solving for $p$. This process will reveal the potential values of $p$ that satisfy the integer constraint. It's like detective work, where we carefully examine each clue to uncover the solution.

Case 1: $4 + p = 1$

Let's start with the smallest factor, 1. If $(4 + p)$ equals 1, we can subtract 4 from both sides to solve for $p$:

$$p = 1 - 4 = -3$$

Uh oh! We hit a snag right away. The value of $p$ we obtained is -3, which is a negative integer. But remember, the problem states that $p$ must be a positive integer. So, this case doesn't work for us. We need to keep searching.

Case 2: $4 + p = 2$

Next up, let's try the factor 2. If $(4 + p)$ equals 2, we subtract 4 from both sides:

$$p = 2 - 4 = -2$$

Another miss! We got $p = -2$, which is also a negative integer. This case doesn't meet our requirement that $p$ be positive. We're still on the hunt.

Case 3: $4 + p = 3$

Moving along to the factor 3, if $(4 + p)$ equals 3, we subtract 4 from both sides:

$$p = 3 - 4 = -1$$

Strike three! We've encountered another negative value for $p$. This case is also a no-go. It's important to remember that in problem-solving, dead ends are just part of the process. We learn from them and keep moving forward.

Case 4: $4 + p = 4$

Let's consider the factor 4. If $(4 + p)$ equals 4, we subtract 4 from both sides:

$$p = 4 - 4 = 0$$

Close, but not quite! We got $p = 0$, which is neither positive nor negative. The problem specifically states that $p$ must be a positive integer, so this case doesn't fit the bill either. We're getting closer, though!

Case 5: $4 + p = 6$

Now, let's try the factor 6. If $(4 + p)$ equals 6, we subtract 4 from both sides:

$$p = 6 - 4 = 2$$

Bingo! We found a winner! $p = 2$ is a positive integer. This value satisfies the condition that $p$ be positive. But let's not stop here. We need to verify that this value of $p$ leads to an integer solution for $x$.

Verifying the Solution

To verify, we plug $p = 2$ back into our equation for $x$:

$$x = \frac{12}{4 + p} = \frac{12}{4 + 2} = \frac{12}{6} = 2$$

Great! When $p = 2$, we get $x = 2$, which is indeed an integer. So, we've successfully found a possible value for $p$ that satisfies the problem's conditions.

Case 6: $4 + p = 12$

Just for completeness, let's examine the last factor, 12. If $(4 + p)$ equals 12, we subtract 4 from both sides:

$$p = 12 - 4 = 8$$

This gives us another positive integer for $p$, $p = 8$. Let's check if it leads to an integer solution for $x$:

$$x = \frac{12}{4 + p} = \frac{12}{4 + 8} = \frac{12}{12} = 1$$

Excellent! When $p = 8$, we get $x = 1$, which is also an integer. So, we've found another possible value for $p$.

The Solution: Unveiling a Possible Value of $p$

After our thorough investigation, we've successfully identified not one, but two possible values for $p$ that lead to integer solutions for $x$:

• p = 2$ (which gives $x = 2$)

• p = 8$ (which gives $x = 1$)

So, a possible value of $p$ is 2 (or 8!).

Conclusion: The Power of Integer Constraints

Guys, this problem beautifully illustrates the power of integer constraints in narrowing down solutions. By recognizing that $x$ had to be an integer, we transformed the problem into a focused search for factors. This approach allowed us to systematically explore possibilities and arrive at the solution. Remember, in math, constraints are your friends! They provide valuable clues and guide you towards the answer.

I hope you enjoyed this mathematical adventure as much as I did. Keep exploring, keep questioning, and keep solving! There's a whole universe of mathematical mysteries waiting to be unlocked.