<Question>
If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then find $x + y + z.$
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find x + y + z given the equations xy = 24, xz = 48, and yz = 72. Hmm, all variables are positive, which is good. I remember that when dealing with systems of equations like this, sometimes multiplying them or dividing them can help simplify things. Let me think step by step.

First, let's list out the equations:

1. xy = 24
2. xz = 48
3. yz = 72

I need to find x, y, z such that all three equations are satisfied, then add them up.

Maybe I can solve for one variable at a time. Let's see. If I can express y and z in terms of x, then perhaps substitute into the third equation. Let's try that.

From equation 1: xy = 24, so y = 24/x.

From equation 2: xz = 48, so z = 48/x.

Then, substitute these into equation 3: yz = 72.

<prompt_1.1> Substitute the expressions for y and z from equations 1 and 2 into equation 3 and simplify. </prompt_1.1>
-----

Alternatively, maybe another approach. Let's see. If I multiply all three equations together: (xy)(xz)(yz) = 24*48*72. That would be x^2 y^2 z^2 = 24*48*72. Then taking the square root of both sides, xyz = sqrt(24*48*72).

<prompt_1.2> Multiply all three equations together and find xyz using the square root of the product. </prompt_1.2>
-----

Alternatively, another way: Let's divide the equations pairwise. For example, equation 2 divided by equation 1: (xz)/(xy) = 48/24 => z/y = 2 => z = 2y. Then, from equation 3: yz =72, and if z=2y, then y*(2y)=72 => 2y^2=72 => y^2=36 => y=6. Then z=12, and from equation1, x=24/y=4. Same result. So, multiple methods lead to the same answer.
</Plan_1>

<Execution_1>
<execution_1.1>
So, y = 24/x and z = 48/x, so yz = (24/x)*(48/x) = (24*48)/(x^2). And this is equal to 72. So:

(24*48)/(x^2) = 72
</execution_1.1>

<execution_1.2>
Let's compute that.

First, compute 24*48*72. Maybe break them down:

24 = 24

48 = 48

72 = 72

Multiply 24*48 first. 24*48: 20*48=960, 4*48=192, so 960+192=1152. Then 1152*72. Hmm, that's a big number. Let me compute step by step.

1152 * 70 = 1152*7*10 = 8064*10 = 80,640

1152 * 2 = 2,304

Adding those together: 80,640 + 2,304 = 82,944

So, x^2 y^2 z^2 = 82,944. Then xyz = sqrt(82,944). Let's compute sqrt(82,944).

Hmm, sqrt(82,944). Let's see. 288^2 = (200 + 88)^2. Wait, 288*288: 200*200=40,000; 200*88=17,600; 88*200=17,600; 88*88=7,744. So, (200 + 88)^2 = 200^2 + 2*200*88 + 88^2 = 40,000 + 35,200 + 7,744 = 82,944. So sqrt(82,944) = 288. Therefore, xyz = 288.

But from equation 1, xy =24, so z = 288 / (xy) = 288 /24 =12. Which matches earlier result. Similarly, xz=48, so y=288 /48=6. And yz=72, so x=288 /72=4. Same results. Therefore, x+y+z=4+6+12=22.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me compute 24*48 first.

<prompt_2.1> Compute 24*48 and solve for x^2. </prompt_2.1>
-----

Based on execution_1.2:
Wait, but maybe I should check if this is correct by plugging back into the original equations.

<prompt_2.2> Verify the solution by checking if x, y, z satisfy the original equations. </prompt_2.2>
-----

Based on execution_1.1:
Let me compute that.

<prompt_2.3> Compute 24*48 and solve for x^2. </prompt_2.3>
-----

Based on execution_1.2:
Alternatively, maybe there's a way to multiply or divide equations to get x + y + z. But I think the straightforward way is solving for each variable.

<prompt_2.4> Discuss the straightforward way to solve for each variable and confirm the solution. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
24*48: 20*48 = 960, 4*48 = 192. Adding those gives 960 + 192 = 1152. So:

1152 / x^2 = 72

Then multiply both sides by x^2:

1152 = 72x^2

Divide both sides by 72:

1152 / 72 = x^2

What's 1152 divided by 72? Let's compute that. 72*16 = 1152, right? 72*10=720, 72*6=432, so 720+432=1152. Yes, so 72*16=1152. Therefore, x^2 = 16. Since x is positive, x = 4.
</execution_2.1>

<execution_2.2>
Let's check:

xy = 4*6 = 24 ✔️

xz = 4*12 = 48 ✔️

yz = 6*12 = 72 ✔️
</execution_2.2>

<execution_2.3>
24*48: Let's see, 20*48 = 960, 4*48 = 192, so 960 + 192 = 1152. So:

1152 / x^2 = 72

Then, multiplying both sides by x^2:

1152 = 72x^2

Divide both sides by 72:

1152 / 72 = x^2

Calculate 1152 ÷ 72. Let's see, 72*16 = 1152 (because 72*10=720, 72*6=432; 720+432=1152). So x^2 = 16. Therefore, x = 4 (since x is positive).
</execution_2.3>

<execution_2.4>
Let me see.

Alternatively, maybe take ratios. For example, equation 2 divided by equation 1: (xz)/(xy) = 48/24 => z/y = 2 => z = 2y. Then, from equation 3: yz =72, substituting z=2y gives y*(2y)=72 => 2y²=72 => y²=36 => y=6. Then z=12. Then from equation 1, x=24/y=24/6=4. So same result. So x=4, y=6, z=12. Thus, x+y+z=22.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now that I have x, I can find y and z.

From equation 1: y = 24/x = 24/4 = 6.

<prompt_3.1> Use x = 4 to find y and z, then calculate x + y + z. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
So y = 6.

From equation 2: z = 48/x = 48/4 = 12. So z = 12.

Let me check these values in equation 3: yz = 6*12 = 72. Which matches the given equation. Great.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So the answer is 4 + 6 + 12 = 22.

Wait, let me double-check all the products to make sure I didn't make a mistake.

<prompt_4.1> Verify all products xy, xz, yz with the found values. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
xy = 4*6 = 24 ✔️

xz = 4*12 = 48 ✔️

yz = 6*12 = 72 ✔️
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the sum of $x$, $y$, and $z$ is $\boxed{22}$.
</Final_answer>
</think>