Linear Equation Word Problems Worksheets - 15 Worksheets Library - Free Printable

Let's solve each of these linear equation word problems step by step. I'll go through each one, set up the equations, and explain the reasoning.

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1. Frederick owns a bakery...

> He sells 50 bags of muffins at $3.50 per bag and 30 muffins at $3 each. How many does he need to sell to make $500?

Let’s define:
- Let $ x $ = number of muffins sold (at $3 each)
- He already sells 50 bags at $3.50 → $ 50 \times 3.50 = 175 $
- So revenue from bags: $175
- Revenue from individual muffins: $ 3x $

Total revenue:
$$
175 + 3x = 500
$$

Solve:
$$
3x = 500 - 175 = 325 \\
x = \frac{325}{3} \approx 108.33
$$

But you can't sell a fraction of a muffin. So he needs to sell 109 muffins to exceed or meet $500.

✔ Answer: 109 muffins

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2. Sarah is planning a rental car...

> Car costs $40/day + $0.25/mile. She plans 300 miles in 4 days. How much will she pay?

Cost = daily rate × days + mileage rate × miles
$$
= 40 \times 4 + 0.25 \times 300 = 160 + 75 = 235
$$

✔ Answer: $235

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3. James works at a bookstore...

> Earns $400 base + $2 per book sold. Earned $300 last month. How many books did he sell?

Wait — he earned $300, but has a base salary of $400? That doesn’t make sense unless the base is not guaranteed.

But it says: "earned $400 with a base salary of $300 plus $2 per book."

So:
- Base = $300
- Commission = $2 per book
- Total earned = $400

Let $ x $ = number of books sold
$$
300 + 2x = 400 \\
2x = 100 \\
x = 50
$$

✔ Answer: 50 books

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4. Farmer wants to build a rectangular pen...

> 400 feet of fencing. Wants to maximize area. What dimensions should he use?

This is a maximization problem using linear equations.

For a rectangle:
- Perimeter: $ P = 2L + 2W = 400 $
- Area: $ A = L \times W $

We want to maximize area given perimeter constraint.

From perimeter:
$$
2L + 2W = 400 \Rightarrow L + W = 200 \Rightarrow W = 200 - L
$$

Area:
$$
A = L(200 - L) = 200L - L^2
$$

This is a quadratic. Maximum occurs at vertex: $ L = 100 $

Then $ W = 100 $

✔ Answer: 100 ft by 100 ft (a square)

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5. Company manufactures laptops...

> Basic: 5 hours assembly, 2 hours testing
> Premium: 8 hours assembly, 3 hours testing
> 400 assembly hours, 150 testing hours available
> How many of each type can be produced?

Let:
- $ x $ = number of basic laptops
- $ y $ = number of premium laptops

Constraints:
- Assembly: $ 5x + 8y \leq 400 $
- Testing: $ 2x + 3y \leq 150 $
- $ x, y \geq 0 $

We want to maximize production, but since no objective function is specified (like profit), assume we want maximum total laptops: $ x + y $

We’ll find feasible integer solutions.

Try solving the system:

Use elimination or substitution.

From testing: $ 2x + 3y \leq 150 $

Let’s try boundary values.

Suppose $ y = 0 $: $ 2x \leq 150 \Rightarrow x \leq 75 $. But check assembly: $ 5(75) = 375 \leq 400 $ → OK. Total: 75

Try $ x = 0 $: $ 3y \leq 150 \Rightarrow y \leq 50 $. Assembly: $ 8×50 = 400 $ → OK. Total: 50

Try mix: suppose $ x = 60 $, then testing: $ 2(60) = 120 $, so $ 3y \leq 30 \Rightarrow y \leq 10 $

Assembly: $ 5(60) + 8(10) = 300 + 80 = 380 \leq 400 $ → OK. Total = 70

Try $ x = 50 $: testing: $ 100 $, so $ 3y \leq 50 \Rightarrow y \leq 16.66 $ → $ y = 16 $

Assembly: $ 5(50) + 8(16) = 250 + 128 = 378 $ → OK. Total = 66

Try $ x = 40 $: testing: $ 80 $, so $ 3y \leq 70 \Rightarrow y \leq 23.3 $ → $ y = 23 $

Assembly: $ 5(40) + 8(23) = 200 + 184 = 384 $ → OK. Total = 63

Now try $ x = 30 $: testing: $ 60 $, $ 3y \leq 90 \Rightarrow y \leq 30 $

Assembly: $ 5(30) + 8(30) = 150 + 240 = 390 $ → OK. Total = 60

Wait — earlier $ x = 60, y = 10 $ gives 70

Can we do better?

Try $ x = 70 $: testing: $ 140 $, so $ 3y \leq 10 \Rightarrow y \leq 3.33 $ → $ y = 3 $

Assembly: $ 5(70) + 8(3) = 350 + 24 = 374 $ → OK. Total = 73 → better!

Try $ x = 72 $: testing: $ 144 $, $ 3y \leq 6 \Rightarrow y = 2 $

Assembly: $ 5(72) + 8(2) = 360 + 16 = 376 $ → OK. Total = 74

Try $ x = 74 $: testing: $ 148 $, $ 3y \leq 2 \Rightarrow y = 0 $

Assembly: $ 5(74) = 370 $ → OK. Total = 74

Try $ x = 73 $: testing: $ 146 $, $ 3y \leq 4 \Rightarrow y = 1 $

Assembly: $ 5(73) + 8(1) = 365 + 8 = 373 $ → OK. Total = 74

Try $ x = 75 $: testing: $ 150 $, $ y = 0 $ → assembly: $ 375 $ → OK. Total = 75

So maximum total is 75 when $ x = 75, y = 0 $

But maybe they want both types? Not stated.

If both must be made, then best is $ x = 73, y = 1 $ → 74

But since not required, max total is 75 basic, 0 premium

But likely, the question wants maximum production under constraints, which is 75 basic laptops.

Wait — but let’s check if we can get more than 75?

No, because $ x = 75 $ uses only 375 assembly hours, and testing is exactly 150.

But can we add some premium?

Try $ x = 70, y = 10 $: testing: $ 2(70) + 3(10) = 140 + 30 = 170 > 150 $ → too much

Try $ x = 60, y = 10 $: testing: $ 120 + 30 = 150 $ → OK

Assembly: $ 5(60) + 8(10) = 300 + 80 = 380 $ → OK

Total: 70

Less than 75.

So maximum total laptops is 75 (all basic).

But perhaps the intent is to maximize output, but not necessarily total units.

Alternatively, maybe they want how many of each can be produced — meaning feasible combinations.

But without an objective, we can't say “optimal”.

But typically in such problems, they expect you to find the maximum possible.

✔ Answer: 75 basic laptops, 0 premium laptops (to maximize total production)

But if both are required, the maximum total is 74 (e.g., 73 basic, 1 premium)

But based on wording: "how many of each type can they produce to maximize profit?" — wait, it says "to maximize profit", but no profit data!

Wait — reread:

> "How many of each type can they produce to maximize profit?"

But no profit per unit is given! So this is incomplete.

Wait — look again:

> "how many of each type can they produce to maximize profit?"

But no profit values are provided. So probably a typo.

Perhaps it's just to find feasible production levels, or maximize total units.

But without profit, we cannot maximize profit.

Alternatively, maybe the question meant "how many of each can they produce?" — i.e., find all possibilities.

But that would be infinite.

So likely, the intended interpretation is: what is the maximum number of laptops they can produce (total), under constraints.

As shown, 75 basic, 0 premium gives 75 laptops.

But let's see if we can get more than 75?

No — because testing limit is 150 hours.

Each laptop needs at least 2 hours testing (basic), so max total laptops = $ 150 / 2 = 75 $

So maximum possible is 75.

And it's achievable with all basic laptops.

✔ Answer: 75 basic laptops, 0 premium laptops

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6. Emily’s savings...

> Invest $5000 in two accounts:
- Account A: 4% interest
- Account B: 6% interest
- Total interest: $260

Let:
- $ x $ = amount in A
- $ 5000 - x $ = amount in B

Interest:
$$
0.04x + 0.06(5000 - x) = 260
$$

Solve:
$$
0.04x + 300 - 0.06x = 260 \\
-0.02x = -40 \\
x = 2000
$$

So:
- $2000 in A
- $3000 in B

✔ Answer: $2000 in Account A, $3000 in Account B

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7. Shipping company charges...

> $10 per package + $2 for every pound over 5 pounds

Let $ w $ = weight in pounds

Cost:
$$
C = 10 + 2(w - 5) = 10 + 2w - 10 = 2w
$$

Wait: $ C = 10 + 2(w - 5) $

Simplify:
$$
C = 10 + 2w - 10 = 2w
$$

So cost = $2w

But only if $ w > 5 $

For $ w \leq 5 $, cost = $10

But the question says: "How much will it cost to ship a package weighing 8 pounds?"

So $ w = 8 $

$$
C = 10 + 2(8 - 5) = 10 + 2(3) = 10 + 6 = 16
$$

✔ Answer: $16

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8. Car rental comparison...

> Plan A: $30/day + $0.30/mile
> Plan B: $20/day + $0.50/mile
> For what mileage are they equal?

Let $ m $ = miles driven

Plan A: $ 30 + 0.30m $
Plan B: $ 20 + 0.50m $

Set equal:
$$
30 + 0.30m = 20 + 0.50m \\
30 - 20 = 0.50m - 0.30m \\
10 = 0.20m \\
m = \frac{10}{0.20} = 50
$$

✔ Answer: 50 miles

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9. Maya bought pens and notebooks...

> 5 pens and 4 notebooks for $30
> Each notebook costs $3 more than a pen

Let:
- $ p $ = price of pen
- $ n $ = price of notebook

Given:
- $ n = p + 3 $
- $ 5p + 4n = 30 $

Substitute:
$$
5p + 4(p + 3) = 30 \\
5p + 4p + 12 = 30 \\
9p = 18 \\
p = 2
$$

Then $ n = 2 + 3 = 5 $

✔ Answer: Pen = $2, Notebook = $5

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10. Train travels from City A to City B...

> Average speed 40 mph
> Round trip total time = 10 hours
> Distance between cities?

Let $ d $ = one-way distance

Time to go: $ \frac{d}{40} $

Time to return: $ \frac{d}{40} $

Total time: $ \frac{d}{40} + \frac{d}{40} = \frac{2d}{40} = \frac{d}{20} $

Set equal to 10:
$$
\frac{d}{20} = 10 \\
d = 200
$$

✔ Answer: 200 miles

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## ✔ Final Answers:

1. 109 muffins
2. $235
3. 50 books
4. 100 ft by 100 ft
5. 75 basic laptops, 0 premium laptops (to maximize total production)
6. $2000 in Account A, $3000 in Account B
7. $16
8. 50 miles
9. Pen: $2, Notebook: $5
10. 200 miles

Let me know if you'd like these formatted as a worksheet answer key!