Okay, so I have this problem where I need to find x + y + z given that xy = 24, xz = 48, and yz = 72. All variables are positive. Hmm, let's see. I need to figure out what x, y, and z are first. Maybe I can solve for one variable in terms of another and substitute. Let me try that.

Starting with the first equation, xy = 24. Maybe I can express x as 24 divided by y. So x = 24/y. Then, substitute this into the second equation xz = 48. Plugging in x, we get (24/y) * z = 48. Let me write that down: (24z)/y = 48. Then, multiply both sides by y to get 24z = 48y. Divide both sides by 24: z = 2y. So z is twice y. That's a relationship between z and y.

Now, let's look at the third equation, yz = 72. We already found that z = 2y, so substitute that into yz: y*(2y) = 72. That simplifies to 2y² = 72. Divide both sides by 2: y² = 36. Take the square root: y = 6 (since y is positive). Then, z = 2y = 2*6 = 12. Now that we have y and z, we can find x from the first equation. x = 24/y = 24/6 = 4. So x is 4, y is 6, z is 12. Then x + y + z = 4 + 6 + 12 = 22. Wait, that seems straightforward. Let me check if all the equations hold.

Check xy: 4*6 = 24 ✔️. Check xz: 4*12 = 48 ✔️. Check yz: 6*12 = 72 ✔️. All equations are satisfied. So the answer should be 22. Hmm, but let me think if there's another way to approach this problem, maybe using ratios or something else, to confirm.

Alternatively, if I multiply all three equations together: (xy)(xz)(yz) = 24*48*72. That would be x²y²z² = (24*48*72). Taking the square root of both sides, xyz = sqrt(24*48*72). Let me compute that. First, multiply 24*48. 24*48 is 1152. Then 1152*72. Let's compute that. 1152*70 = 80,640 and 1152*2 = 2,304. So total is 80,640 + 2,304 = 82,944. So xyz = sqrt(82,944). Let me find the square root of 82,944. 288 squared is 82,944 because 288*288: 200*200=40,000; 200*88=17,600; 88*200=17,600; 88*88=7,744. Wait, that's not the right way. Alternatively, 288*288: 200^2 + 2*200*88 + 88^2 = 40,000 + 35,200 + 7,744 = 82,944. Yes, so sqrt(82,944) = 288. Therefore, xyz = 288.

Now, since we already found that x=4, y=6, z=12, their product is 4*6*12=288, which matches. So that's consistent. But maybe we can use this to find another way. If xyz=288, and we know from the first equation that xy=24, so z = xyz / xy = 288 /24 =12. Which matches our earlier result. Similarly, x = xyz / yz =288 /72=4, and y = xyz / xz =288 /48=6. So that's another way to find each variable. But since the problem only asks for x + y + z, once we have all three variables, adding them is straightforward.

Alternatively, maybe there's a way to find x + y + z without finding each variable individually. Let's see. If we have the three products: xy=24, xz=48, yz=72. Let me see if there's a way to express x + y + z in terms of these. Hmm, not directly obvious. Maybe we can think about ratios. For example, xz = 48 and xy=24, so xz / xy = 48/24 = 2. That simplifies to z/y = 2, so z=2y, which is the same as before. Then yz=72, so substituting z=2y gives 2y²=72, so y²=36, y=6. So same result. So maybe this is the most straightforward way.

But just to make sure I didn't miss anything, let me check my steps again. Starting with xy=24, so x=24/y. Then xz=48: (24/y)*z=48 →24z=48y → z=2y. Then yz=72: y*(2y)=72 →2y²=72 →y²=36 →y=6. Then z=12, x=4. Adding up gives 4+6+12=22. All equations check out. So I think that's correct. I don't see any mistakes here. Maybe another approach: suppose we let a = x, b = y, c = z. Then ab=24, ac=48, bc=72. Then, solving for a, b, c. Let's see, from ab=24 and ac=48, divide the second equation by the first: (ac)/(ab) = 48/24 = 2 → c/b = 2 → c=2b. Then bc=72: b*(2b)=2b²=72 →b²=36→b=6. Then c=12, and a=24/b=24/6=4. Same result. So same steps, just different variable names. So that's consistent.

Alternatively, suppose we take the product of all three equations: (ab)(ac)(bc)= (24)(48)(72). As before, this is a²b²c²=24*48*72. Taking square roots: abc=288. Then, since ab=24, then c=288/24=12. Similarly, ac=48, so b=288/48=6. And bc=72, so a=288/72=4. So same answer. So this method also works. So regardless of the method used, the answer is 22. I think that's solid. So I can be confident the answer is 22.
Therefore, the sum of $x$, $y$, and $z$ is $\boxed{22}$.
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