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

Here is the complete solution for all six word problems, including writing the inequality, solving it, and explaining the reasoning. Since I cannot graph here, I will describe how to graph each solution on a number line.

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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:
Total weight ≤ 1000
→ \( 300 + 25x \leq 1000 \)

Solve:
Subtract 300 from both sides:
\( 25x \leq 700 \)

Divide by 25:
\( x \leq 28 \)

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

Graph:
On a number line, draw a closed circle at 28 (since ≤ includes 28), and shade everything to the left (toward negative infinity).

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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:
\( n \) = number of nights

Inequality:
Cost = 100n + 35 ≤ 500

Solve:
Subtract 35:
\( 100n \leq 465 \)

Divide by 100:
\( n \leq 4.65 \)

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:
Closed circle at 4, shade to the 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:
\( j \) = total number of juniors

Inequality:
“no less than a third” means ≥ one-third
→ \( \frac{1}{3}j \leq 37 \) ✘ — Wait! That’s backwards.

Actually:
If 37 juniors went, and that is *at least* one-third of all juniors, then:

→ \( 37 \geq \frac{1}{3}j \)
or equivalently:
→ \( \frac{1}{3}j \leq 37 \) — still wrong logic.

✔ Correct interpretation:
“At least one-third went” → Number who went ≥ one-third of total
So:
\( 37 \geq \frac{1}{3}j \)

Multiply both sides by 3:
\( 111 \geq j \) → \( j \leq 111 \) ✘ — again, wrong direction.

Wait — let’s rephrase:

If “no less than a third went”, that means the 37 juniors represent at least one-third of the total. So:

\( 37 \geq \frac{1}{3}j \)

But this gives \( j \leq 111 \), which would mean if only 111 juniors exist, then 37 is more than a third? Let’s check:

One-third of 111 is 37 → so 37 is exactly one-third.

The phrase “no less than a third” means ≥ one-third → so 37 ≥ (1/3)j → j ≤ 111.

BUT — if there are fewer than 111 juniors, say 100, then one-third is ~33.3, and 37 > 33.3 → satisfies “no less than a third”.

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

That is — what is the smallest possible number of juniors such that 37 is at least one-third?

If j = 111 → 1/3 of 111 = 37 → exactly one-third → satisfies “no less than”.

If j = 110 → 1/3 of 110 ≈ 36.67 → 37 > 36.67 → still satisfies.

Wait — actually, as j decreases, 1/3 j decreases, so 37 becomes larger relative to 1/3 j — so any j ≤ 111 works? But that gives an upper bound.

We need the minimum number of juniors — meaning the smallest possible total that still allows 37 to be ≥ one-third.

To make 37 ≥ (1/3)j, solve for j:

Multiply both sides by 3:
\( 111 \geq j \) → \( j \leq 111 \)

But that gives an upper limit — not a lower limit.

I think I misread.

Let me re-express:

“If no less than a third of the juniors went to prom” — meaning the number who went is ≥ (1/3) of total juniors.

Given: 37 juniors went → so:

\( 37 \geq \frac{1}{3}j \)

Multiply both sides by 3:

\( 111 \geq j \) → \( j \leq 111 \)

This suggests the total number of juniors must be at most 111.

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

That doesn’t make sense — if j can be as small as 1, then 37 is way more than one-third — so technically, any j ≤ 111 satisfies the condition.

But that can’t be right — because if there are only 10 juniors, you can’t have 37 go to prom.

Ah! Crucial point: You can’t have more juniors going than exist.

So, logically, the number of juniors who went (37) cannot exceed the total number of juniors.

So we also have:
\( j \geq 37 \)

And from the condition:
\( 37 \geq \frac{1}{3}j \) → \( j \leq 111 \)

So combining:
\( 37 \leq j \leq 111 \)

Now, the question asks for the minimum amount of juniors — so the smallest possible j that satisfies both.

That is: j = 37

Check: If there are 37 juniors, then one-third is about 12.3 — and 37 ≥ 12.3 → true.

Also, 37 juniors went — which is possible if total is 37.

✔ So minimum number of juniors is 37.

Wait — but if only 37 juniors exist, and all 37 went, that’s 100% — which is certainly “no less than a third”. So yes.

But is there a smaller number? No — because you can’t have 37 juniors going if there are fewer than 37 juniors.

So the constraint \( j \geq 37 \) is necessary.

Thus, minimum j is 37.

Inequality setup:

We have two constraints:
1. \( j \geq 37 \) (can’t have more going than total)
2. \( 37 \geq \frac{1}{3}j \) → \( j \leq 111 \)

But since we want the minimum j, we pick the smallest possible — which is 37.

✔ Answer: Minimum number of juniors is 37.

Graph:
Closed circle at 37, shade to the right? No — wait, we’re looking for minimum value, but the solution set is j ≥ 37 and j ≤ 111 → so [37, 111]. But since the question asks for the *minimum*, the answer is 37.

For graphing the solution set: closed circles at 37 and 111, shade in between.

But since the problem asks for the minimum amount, we report 37.

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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:
\( s \) = calories per smoothie

Inequality:
Total calories < 2000
→ \( 875 + 3s < 2000 \)

Solve:
Subtract 875:
\( 3s < 1125 \)

Divide by 3:
\( s < 375 \)

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

Graph:
Open circle at 375 (since <, not ≤), 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:
“2 less than 3 times a number” → \( 3x - 2 \)
“is greater than 10” → \( 3x - 2 > 10 \)

Solve:
Add 2:
\( 3x > 12 \)

Divide by 3:
\( x > 4 \)

✔ Answer: The number is greater than 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:
\( b \) = brother’s age
Then Ben’s age = \( 2b \)

Inequality:
Sum of ages < 81
→ \( b + 2b < 81 \)
→ \( 3b < 81 \)

Solve:
Divide by 3:
\( b < 27 \)

✔ Answer: Ben’s brother could be any age less than 27 (e.g., 26, 25, etc.).

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

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## ✔ Final Answers Summary:

1. 28 boxes
2. 4 nights
3. 37 juniors *(minimum)*
4. Each smoothie < 375 calories
5. Number > 4
6. Brother’s age < 27

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## 📊 Graphing Instructions (for each):

- Use a horizontal number line.
- For ≤ or ≥: use a closed circle.
- For < or >: use an open circle.
- Shade the region that satisfies the inequality.

Example for Problem 1 (x ≤ 28):
Draw line, mark 28 with closed circle, shade left.

Example for Problem 5 (x > 4):
Mark 4 with open circle, shade right.

You can now fill in the worksheet with these answers and graphs!

Let me know if you’d like visual sketches of the graphs.