What’s the real trick behind “PQ = 6x + 25, QR = 16 – 3x, find PR”?
You’ve probably seen that line of algebra‑geometry mash‑up in a homework sheet or a test prep book and thought, “Great, another word problem that looks more like a puzzle than a math question.” The short answer is: it’s a simple application of the triangle inequality, plus a dash of algebra, and a sprinkle of intuition about what the sides of a triangle can actually be.
Below you’ll get the whole story—what the symbols mean, why the answer matters, the step‑by‑step walk‑through, common slip‑ups, and a handful of tips you can reuse on any similar problem Simple, but easy to overlook..
What Is This Problem, Anyway?
In plain English, the problem gives you two sides of a triangle expressed as linear expressions in the same unknown x:
- PQ = 6x + 25
- QR = 16 – 3x
Your job is to figure out the length of the third side, PR, without any extra information like angles or area.
That sounds impossible at first—how can you find a single number when you have an unknown variable floating around? The secret is that the three lengths must obey the triangle inequality. Practically speaking, in other words, any two sides added together have to be strictly longer than the third side. Those three simple rules are the only constraints we have, and they’re enough to pin down x, which then lets us compute PR.
The triangle inequality in a nutshell
For any triangle with sides a, b, c:
- a + b > c
- a + c > b
- b + c > a
If you plug the expressions for PQ and QR into those three inequalities, you’ll end up with a range of possible x‑values. The only value that satisfies all three will give you a valid triangle, and that’s the x you need.
No fluff here — just what actually works The details matter here..
Why It Matters
You might wonder, “Why bother with this algebraic gymnastics? I could just measure the sides on a drawing.”
- Real‑world design – Engineers often know relationships between dimensions (e.g., a beam that’s 6 × x + 25 cm long) but not the exact numbers. The triangle inequality tells them whether a design is even physically possible.
- Test prep – Standardized tests love to hide geometry behind algebra. Mastering this trick saves you minutes and points.
- Critical thinking – It forces you to look at constraints, not just equations. That habit pays off in programming, finance, and everyday problem solving.
How It Works (Step‑by‑Step)
Let’s break the process into bite‑size pieces.
1. Write down the three inequalities
We have three sides:
- a = PQ = 6x + 25
- b = QR = 16 – 3x
- c = PR = ?
The triangle inequality gives us:
- (6x + 25) + (16 – 3x) > PR
- (6x + 25) + PR > (16 – 3x)
- (16 – 3x) + PR > (6x + 25)
We don’t know PR yet, but we do know that PR must also be a positive length. The trick is to express PR in terms of x using the fact that the three sides must satisfy the inequality as tightly as possible. In practice, the only way to nail down a single value for PR is to assume the triangle is degenerate—the case where the sum of the two known sides equals the third. That gives us the boundary condition we need Most people skip this — try not to..
2. Set up the “degenerate” equation
If you add the two known sides, you get a number that could be the maximum possible length for PR. Anything longer would break the inequality, anything shorter would still be okay, but the problem usually expects the exact length that makes the triangle just possible.
Counterintuitive, but true.
So we write:
[ PR = (6x + 25) + (16 - 3x) = 6x + 25 + 16 - 3x = 3x + 41 ]
Now PR is also expressed in terms of x. Great—now every side is a linear function of x.
3. Plug PR back into the other two inequalities
We still have to make sure the other two triangle‑inequality conditions hold:
- (6x + 25) + (3x + 41) > (16 – 3x)
- (16 – 3x) + (3x + 41) > (6x + 25)
Let’s simplify each Simple as that..
Inequality A
[
6x + 25 + 3x + 41 > 16 - 3x
\Rightarrow 9x + 66 > 16 - 3x
\Rightarrow 12x > -50
\Rightarrow x > -\frac{50}{12} \approx -4.17
]
Since a length can’t be negative, any positive x will satisfy this one automatically.
Inequality B
[
16 - 3x + 3x + 41 > 6x + 25
\Rightarrow 57 > 6x + 25
\Rightarrow 32 > 6x
\Rightarrow x < \frac{32}{6} \approx 5.33
]
Now we have a real restriction: x must be less than about 5.33 Not complicated — just consistent. That's the whole idea..
4. Combine the restrictions
From step 3 we have:
- x > ‑4.17 (practically x > 0)
- x < 5.33
So any x in the interval 0 < x < 5.33 will give a valid triangle. The problem, however, usually expects a single integer value for x that makes the side lengths whole numbers. The most natural choice is the largest integer that fits the range: x = 5.
5. Compute the sides with x = 5
- PQ = 6·5 + 25 = 30 + 25 = 55
- QR = 16 – 3·5 = 16 – 15 = 1
- PR = 3·5 + 41 = 15 + 41 = 56
Check the triangle inequality quickly:
- 55 + 1 = 56 → equals PR (degenerate case)
- 55 + 56 = 111 > 1 → ok
- 1 + 56 = 57 > 55 → ok
So the triangle is right on the edge of collapsing into a straight line, which is exactly what our “degenerate” assumption gave us And that's really what it comes down to..
Answer: PR = 56 (when x = 5) Small thing, real impact..
If the problem didn’t ask for an integer, you could leave PR as 3x + 41 and say “any x between 0 and 5.33 works; PR will be 3x + 41.”
Common Mistakes / What Most People Get Wrong
-
Forgetting the degenerate case – Many students try to solve for PR directly from the three inequalities and end up with a system that seems impossible. The key is to realize that the only way to lock down a single value is to treat the sum of the known sides as the maximum PR can be And that's really what it comes down to..
-
Dropping the “>” sign – The triangle inequality is strict; “≥” would let a flat line count as a triangle, which most textbooks disallow. Using “≥” can give you an extra range of x that isn’t actually valid But it adds up..
-
Ignoring the positivity of lengths – It’s easy to get a negative bound like x > ‑4.17 and think you’re done. Remember that side lengths must be positive, so you always intersect the algebraic range with x > 0.
-
Assuming x must be an integer – The problem statement didn’t say “x is an integer,” but many test‑prep guides do. If you’re not forced to pick an integer, you can leave the answer in terms of x.
-
Mixing up which side is which – When you write the three inequalities, it’s tempting to swap sides around. Stick to a consistent labeling (PQ, QR, PR) and you’ll avoid a lot of needless algebra But it adds up..
Practical Tips / What Actually Works
-
Write the degenerate equation first. As soon as you see two sides given in terms of the same variable, add them together and set that equal to the third side. It gives you a clean expression for the unknown side And that's really what it comes down to..
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Sketch it. Even a quick doodle of a triangle with the three expressions labeled helps you see which inequality will be the tightest Easy to understand, harder to ignore. Practical, not theoretical..
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Check endpoints. After you solve the inequalities, test the boundary values (e.g., x = 5.33) to see if they make any side zero or negative. Those are the “no‑go” points Took long enough..
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Keep an eye on units. If the problem mentions centimeters or inches elsewhere, keep that unit in mind; it can catch a slip where you accidentally treat a length as a pure number.
-
Use a calculator for fractions. Converting (-50/12) to (-4.17) isn’t necessary for the final answer, but it helps you quickly see that the lower bound is irrelevant once you enforce positivity.
FAQ
Q1: Do I always have to assume a degenerate triangle?
A: Not always, but it’s the fastest way to get a single value for the unknown side when only two sides are expressed in terms of the same variable. If the problem gives an angle or area, you’d use a different method.
Q2: What if the resulting x isn’t an integer?
A: Then you leave PR as the expression you derived (here, 3x + 41) and state the valid range for x. Many textbooks prefer a whole‑number answer, so they’ll pick the nearest integer that keeps the triangle valid.
Q3: Can the triangle inequality ever be an equality?
A: Only in the degenerate case, where the three points lie on a straight line. Most geometry courses treat that as “not a triangle,” but it’s useful for boundary analysis Easy to understand, harder to ignore..
Q4: Why is the upper bound x < 5.33 and not x ≤ 5.33?
A: Because the inequality 57 > 6x + 25 is strict (>), not ≥. If x were exactly 5.33, the left side would equal the right side, breaking the rule.
Q5: Could QR become negative for some x?
A: Yes—if x > 16/3 ≈ 5.33, QR = 16 – 3x would dip below zero, which is impossible for a length. That’s another reason why the upper bound matters.
That’s the whole picture. You’ve seen why the triangle inequality is the hidden hero, walked through the algebra, spotted the usual pitfalls, and walked away with a concrete answer: PR = 56 (when x = 5).
Next time you spot “6x + 25, 16 – 3x, find the third side,” you’ll know exactly what to do—no panic, just a quick sketch, a couple of inequalities, and you’re done. Happy problem‑solving!
