One Variable Inequalities Notes and Worksheets - Lindsay Bowden - Free Printable

Here is the complete solution for all six one-variable inequality word problems, including writing the inequality, solving it, and graphing the answer.

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Problem 1:
> A boat can hold at most 1,000 pounds. If there is 300 pounds of equipment plus 25-pound boxes, what is the maximum number of boxes the ship can carry?

Let x = number of 25-pound boxes

Inequality:
> 300 + 25x ≤ 1000

Solve:
> 25x ≤ 700
> x ≤ 28

✔ Answer: The boat can carry at most 28 boxes.

Graph: On a number line, shade from negative infinity to 28 (inclusive). Place a closed circle at 28 and shade left.

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Problem 2:
> A hotel costs $100 per night plus a $35 hotel fee. If George has $500, what is the maximum number of nights he could stay?

Let x = number of nights

Inequality:
> 100x + 35 ≤ 500

Solve:
> 100x ≤ 465
> x ≤ 4.65

Since George can’t stay a fraction of a night, we take the greatest whole number less than or equal to 4.65, which is 4.

✔ Answer: George can stay at most 4 nights.

Graph: Shade from negative infinity to 4 (inclusive). Closed circle at 4, shade left.

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Problem 3:
> The school paper says that no less than a third of the juniors went to prom. If 37 juniors went, what’s the minimum amount of juniors at the school?

Let x = total number of juniors

“No less than a third” means ≥ 1/3.

Inequality:
> (1/3)x ≤ 37 ← Wait! This is backwards.

Actually: “No less than a third went” → the number who went (37) is at least a third of total juniors.

So:
> 37 ≥ (1/3)x
or equivalently:
> (1/3)x ≤ 37 → No, that’s incorrect.

Correct interpretation:
> 37 ≥ (1/3)x → Multiply both sides by 3:
> 111 ≥ x → x ≤ 111? That doesn’t make sense — if 37 is *at least* a third, then total must be *at most* 111? But that contradicts “minimum amount”.

Let’s rephrase:

If 37 juniors went, and that is no less than a third of all juniors, then:

> 37 ≥ (1/3)x

Multiply both sides by 3:
> 111 ≥ x → x ≤ 111

But this gives the maximum possible juniors, not minimum.

Wait — let’s think logically.

If 37 is at least one-third of the total, then the total must be at most 111? That doesn’t give us the minimum.

Actually, the phrase “no less than a third” means the proportion who went is ≥ 1/3.

So:
> 37 / x ≥ 1/3

Now solve:

Multiply both sides by x (assuming x > 0):
> 37 ≥ (1/3)x

Then multiply both sides by 3:
> 111 ≥ x → x ≤ 111

Still same result.

But the question asks: “what’s the minimum amount of juniors at the school?”

If 37 is at least a third, then the smallest possible total number of juniors occurs when 37 is exactly a third.

So:
> (1/3)x = 37 → x = 111

If there were fewer than 111 juniors, say 110, then a third would be ~36.67, and 37 > 36.67 — so 37 is still more than a third. So 110 would work.

Wait — actually, if x = 110, then 1/3 of 110 ≈ 36.67, and 37 > 36.67 → satisfies “no less than a third”.

If x = 109, 1/3 ≈ 36.33, 37 > 36.33 → still works.

What’s the minimum x such that 37 ≥ (1/3)x?

As x gets smaller, (1/3)x gets smaller, so 37 is even larger than a third → still satisfies.

But that suggests there’s no lower bound? That can’t be right.

Ah — I see the confusion.

The problem says: “no less than a third of the juniors went to prom.” — meaning, the number who went (37) is ≥ (1/3) of total juniors.

So:
> 37 ≥ (1/3)x
→ x ≤ 111

This tells us the maximum number of juniors is 111.

But the question asks for the minimum number of juniors.

That doesn’t make sense unless we misinterpreted.

Actually — if 37 juniors went, and that represents at least a third of the total, then the total number of juniors cannot exceed 111. But it could be less — as long as 37 is still ≥ 1/3 of that smaller number.

For example, if total juniors = 50, then 1/3 of 50 ≈ 16.67, and 37 > 16.67 → satisfies.

If total juniors = 10, 1/3 ≈ 3.33, 37 > 3.33 → still satisfies.

So technically, there is no minimum — any number of juniors ≤ 111 works, but since 37 went, the total must be at least 37 (you can’t have more people go than exist).

Ah! That’s the key.

The total number of juniors must be at least 37 (since 37 went), and also must satisfy 37 ≥ (1/3)x → x ≤ 111.

So the total number of juniors x must satisfy:

> 37 ≤ x ≤ 111

Therefore, the minimum number of juniors is 37.

But does 37 satisfy “no less than a third”? If total juniors = 37, then a third is ≈12.33, and 37 ≥ 12.33 → yes.

So the minimum is 37.

✔ Answer: The minimum number of juniors at the school is 37.

Graph: Shade from 37 to 111 inclusive. Closed circles at both ends.

Wait — but the question only asks for the minimum, so perhaps just x ≥ 37? But the constraint is 37 ≥ (1/3)x, which gives x ≤ 111, and x ≥ 37 (since you can't have fewer than 37 juniors if 37 went).

Actually, the problem is likely expecting us to solve:

> 37 ≥ (1/3)x → x ≤ 111

But that gives maximum, not minimum.

I think there’s a wording issue.

Standard interpretation in math problems: “no less than a third went” with 37 going → find minimum total.

Actually, the minimum total occurs when 37 is exactly one-third — because if total is larger, 37 is less than a third? No.

Let me flip it.

Suppose total juniors = T.

We are told: number who went ≥ (1/3)T → 37 ≥ (1/3)T → T ≤ 111.

Also, T ≥ 37 (since 37 juniors went).

So T ∈ [37, 111]

Thus, minimum T is 37.

✔ Final Answer: 37

Graph: Number line shaded from 37 to 111, closed circles at both ends.

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Problem 4:
> Kyle’s doctor told him that in order to lose weight, he needs to eat less than 2,000 calories a day. He eats 875 calories for breakfast and then 3 protein smoothies. How many calories can the protein smoothies have to reach his goal?

Let x = calories per smoothie

Total calories: 875 + 3x < 2000

Solve:
> 3x < 1125
> x < 375

✔ Answer: Each protein smoothie must have less than 375 calories.

Graph: Open circle at 375, shade to the left.

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Problem 5:
> 2 less than 3 times a number is greater than 10.

Let x = the number

Inequality:
> 3x - 2 > 10

Solve:
> 3x > 12
> x > 4

✔ Answer: x > 4

Graph: Open circle at 4, shade to the right.

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Problem 6:
> Ben is twice his brother’s age. If the sum of their ages is below 81, how old could Ben’s brother be?

Let x = brother’s age

Then Ben’s age = 2x

Sum: x + 2x < 81 → 3x < 81 → x < 27

✔ Answer: Ben’s brother could be any age less than 27 years old.

Graph: Open circle at 27, shade to the left.

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## ✔ Final Answers Summary:

1. x ≤ 28 → Max 28 boxes
2. x ≤ 4.65 → x ≤ 4 (whole nights) → Max 4 nights
3. 37 ≤ x ≤ 111 → Min 37 juniors
4. x < 375 → Each smoothie < 375 cal
5. x > 4
6. x < 27 → Brother < 27 years old

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## 📈 Graphing Instructions (for each):

- Use a number line.
- For ≤ or ≥: closed circle
- For < or >: open circle
- Shade in the direction of the inequality.

Example for Problem 1 (x ≤ 28):
```
<---|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|---->
... 25 26 27 28 29 30 ...
●================>
```

(Shade left from 28, closed circle)

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All problems solved with inequalities, solutions, and graphing guidance.