Linear Inequalities Worksheets with Answer Key - Free Printable

Let's solve each problem step by step, including writing inequalities, interpreting them, and explaining how to graph the system. Since I can't see the image directly, I'll go through both word problems based on your description.

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Problem 1: Sara is buying wings and hot dogs

Given:
- One package of wings costs $7.
- Hot dogs cost $4 per pound.
- Total spending must be less than $40.
- She buys at least 5 pounds of hot dogs.

Let:
- $ x $ = number of packages of wings
- $ y $ = number of pounds of hot dogs

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#### a) Write an inequality representing the cost.

Total cost = (cost of wings) + (cost of hot dogs)

$$
7x + 4y < 40
$$

This represents that the total cost must be less than $40.

✔ Answer: $ 7x + 4y < 40 $

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#### b) Write an inequality showing the condition if Sara buys at least 5 pounds of hot dogs.

"At least 5 pounds" means:

$$
y \geq 5
$$

✔ Answer: $ y \geq 5 $

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#### c) Graph the system of inequalities and find how many packages of wings and hot dogs she can buy at most?

We now have the system:

1. $ 7x + 4y < 40 $
2. $ y \geq 5 $

We’ll graph these and find feasible integer solutions (since you can’t buy a fraction of a package or pound in real life).

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##### Step 1: Graph $ 7x + 4y < 40 $

First, graph the boundary line: $ 7x + 4y = 40 $

Find intercepts:

- When $ x = 0 $: $ 4y = 40 $ → $ y = 10 $
- When $ y = 0 $: $ 7x = 40 $ → $ x \approx 5.71 $

So the line goes from (0,10) to (5.71, 0)

Since it's <, shade below the line (but not including the line).

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##### Step 2: Graph $ y \geq 5 $

Draw horizontal line at $ y = 5 $, and shade above it (including the line).

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##### Step 3: Find the overlapping region (feasible region)

The feasible region is where both conditions are true:
- Below the line $ 7x + 4y = 40 $
- Above or on $ y = 5 $

Now we want to find how many packages of wings and hot dogs she can buy at most — likely meaning the maximum values of $ x $ and $ y $ within this region.

But since it's "at most", we interpret it as: what are the maximum possible values of $ x $ and $ y $ such that both constraints are satisfied?

Let’s find integer points in the feasible region.

We know $ y \geq 5 $, so try values of $ y $ starting from 5 upward.

Plug into the first inequality: $ 7x + 4y < 40 $

Try $ y = 5 $:
- $ 7x + 4(5) < 40 $ → $ 7x + 20 < 40 $ → $ 7x < 20 $ → $ x < 2.857 $
→ So max $ x = 2 $

Try $ y = 6 $:
- $ 7x + 24 < 40 $ → $ 7x < 16 $ → $ x < 2.285 $ → $ x = 2 $

Try $ y = 7 $:
- $ 7x + 28 < 40 $ → $ 7x < 12 $ → $ x < 1.71 $ → $ x = 1 $

Try $ y = 8 $:
- $ 7x + 32 < 40 $ → $ 7x < 8 $ → $ x < 1.14 $ → $ x = 1 $

Try $ y = 9 $:
- $ 7x + 36 < 40 $ → $ 7x < 4 $ → $ x < 0.57 $ → $ x = 0 $

Try $ y = 10 $:
- $ 7x + 40 < 40 $ → $ 7x < 0 $ → $ x < 0 $ → not possible

So maximum values:

- Maximum $ y $: 9 (with $ x = 0 $)
- Maximum $ x $: 2 (with $ y = 5 $ or $ y = 6 $)

But the question says: "how many packages of wings and how many packages of hotdogs can she buy at most?"

It might be asking for the maximum possible combinations.

But since both can't be maximized simultaneously, we interpret it as:

> What is the maximum number of packages of wings she can buy?
> What is the maximum number of pounds of hot dogs she can buy?

So:

- Max wings: 2 packages (when $ y = 5 $ or $ y = 6 $)
- Max hot dogs: 9 pounds (when $ x = 0 $)

✔ Answer:
- At most, Sara can buy 2 packages of wings and 9 pounds of hot dogs (not both at the same time).
- But if she wants to maximize one, then:
- Max wings: 2
- Max hot dogs: 9

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✔ Final Answer for Problem 1:

a) $ 7x + 4y < 40 $
b) $ y \geq 5 $
c) Graph the two inequalities:
- Shade below $ 7x + 4y = 40 $ (dashed line)
- Shade above $ y = 5 $ (solid line)
- Feasible region is intersection
- Maximum wings: 2 packages
- Maximum hot dogs: 9 pounds

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Problem 2: Driving with friend

Given:
- You drive at 70 mph
- Friend drives at 60 mph
- Together, you want to drive:
- Less than 15 hours per day
- At least 600 miles per day
- Your friend drives longer than you

Let:
- $ x $ = number of hours you drive
- $ y $ = number of hours friend drives

Constraints:

1. Total time: $ x + y < 15 $
2. Total distance: $ 70x + 60y \geq 600 $
3. Friend drives longer: $ y > x $

We need to find possible values of $ x $ and $ y $ satisfying all three.

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#### Step 1: Inequalities

1. $ x + y < 15 $
2. $ 70x + 60y \geq 600 $
3. $ y > x $

We’ll graph and find feasible region.

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##### Graphing:

1. $ x + y < 15 $ → boundary line $ x + y = 15 $, dashed, shade below
2. $ 70x + 60y \geq 600 $ → divide by 10: $ 7x + 6y \geq 60 $
- Boundary: $ 7x + 6y = 60 $
- Intercepts:
- $ x = 0 $: $ 6y = 60 $ → $ y = 10 $
- $ y = 0 $: $ 7x = 60 $ → $ x \approx 8.57 $
- Shade above the line
3. $ y > x $ → shade above the line $ y = x $

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##### Find feasible region:

We want the area where:
- Below $ x + y = 15 $
- Above $ 7x + 6y = 60 $
- Above $ y = x $

Let’s find intersection points.

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Intersection of $ x + y = 15 $ and $ 7x + 6y = 60 $:

From first: $ y = 15 - x $

Substitute into second:

$$
7x + 6(15 - x) = 60 \\
7x + 90 - 6x = 60 \\
x + 90 = 60 \\
x = -30
$$

Not valid (negative time). So no intersection in feasible region.

Try $ 7x + 6y = 60 $ and $ y = x $:

$$
7x + 6x = 60 → 13x = 60 → x ≈ 4.615, y ≈ 4.615
$$

Check if $ x + y < 15 $? $ 4.615 + 4.615 = 9.23 < 15 $ → yes

Now check $ x + y = 15 $ and $ y = x $:

$ x + x = 15 → x = 7.5, y = 7.5 $

But $ y > x $ is not satisfied (equal), so not included.

So feasible region is bounded by:

- $ y > x $
- $ x + y < 15 $
- $ 7x + 6y \geq 60 $

Let’s find where $ 7x + 6y = 60 $ intersects $ x + y = 15 $

Wait earlier we got negative value — let’s double-check:

From $ x + y = 15 $ → $ y = 15 - x $

Plug into $ 7x + 6y = 60 $:

$$
7x + 6(15 - x) = 60 \\
7x + 90 - 6x = 60 \\
x + 90 = 60 \\
x = -30 → invalid
$$

So lines don’t intersect in positive quadrant.

Now, $ 7x + 6y = 60 $ intersects $ x + y = 15 $ only at negative values, so the constraint $ 7x + 6y \geq 60 $ is binding near origin.

Let’s find where $ 7x + 6y = 60 $ crosses $ x = 0 $: $ y = 10 $

And $ y = 0 $: $ x = 60/7 ≈ 8.57 $

So the line $ 7x + 6y = 60 $ starts at (0,10) and goes to (8.57, 0)

Now, we want $ x + y < 15 $ → which is almost always true if $ x, y $ are less than 15

But also $ y > x $

Let’s find minimum possible values satisfying all.

Try to find possible values of $ x $ and $ y $

Let’s suppose $ x = 4 $, then $ y > 4 $, and $ x + y < 15 $ → $ y < 11 $

Distance: $ 70(4) + 60y = 280 + 60y \geq 600 $

→ $ 60y \geq 320 $ → $ y \geq 5.333 $

So $ y \geq 5.333 $, $ y > 4 $, $ y < 11 $

So $ y \in [5.333, 11) $

Pick $ y = 6 $: total time = 4+6=10 < 15, distance = 280 + 360 = 640 ≥ 600 → OK

Try $ x = 5 $, $ y > 5 $, $ y < 10 $

Distance: $ 70(5) + 60y = 350 + 60y \geq 600 $ → $ 60y \geq 250 $ → $ y \geq 4.166 $, but already $ y > 5 $

So $ y \in (5, 10) $

Say $ y = 6 $: total time = 11 < 15, distance = 350 + 360 = 710 → OK

Can we go higher?

Try $ x = 6 $, $ y > 6 $, $ x + y < 15 $ → $ y < 9 $

Distance: $ 70(6) = 420 $, need $ 60y \geq 180 $ → $ y \geq 3 $ → already satisfied

So $ y \in (6, 9) $

Say $ y = 7 $: total time = 13 < 15, distance = 420 + 420 = 840 → OK

Try $ x = 7 $, $ y > 7 $, $ x + y < 15 $ → $ y < 8 $

So $ y \in (7, 8) $

Distance: $ 70(7) = 490 $, need $ 60y \geq 110 $ → $ y \geq 1.83 $ → OK

So $ y = 7.5 $: total time = 14.5 < 15, distance = 490 + 450 = 940 → OK

Try $ x = 8 $, $ y > 8 $, $ x + y < 15 $ → $ y < 7 $ → contradiction

So $ x $ cannot be 8 or more.

Thus, maximum $ x = 7 $, $ y $ between 7 and 8

Maximum $ y $: when $ x $ is small

Try $ x = 3 $, $ y > 3 $, $ x + y < 15 $ → $ y < 12 $

Distance: $ 70(3) = 210 $, need $ 60y \geq 390 $ → $ y \geq 6.5 $

So $ y \in (6.5, 12) $

Say $ y = 10 $: total time = 13 < 15, distance = 210 + 600 = 810 → OK

Try $ x = 2 $, $ y > 2 $, $ y < 13 $

Distance: $ 140 + 60y \geq 600 $ → $ 60y \geq 460 $ → $ y \geq 7.666 $

So $ y \in (7.666, 13) $

Say $ y = 12 $: total time = 14 < 15, distance = 140 + 720 = 860 → OK

Try $ x = 1 $, $ y > 1 $, $ y < 14 $

Distance: $ 70 + 60y \geq 600 $ → $ 60y \geq 530 $ → $ y \geq 8.833 $

So $ y \in (8.833, 14) $

Say $ y = 13 $: total time = 14 < 15, distance = 70 + 780 = 850 → OK

Try $ x = 0 $: $ y > 0 $, $ y < 15 $, distance = $ 60y \geq 600 $ → $ y \geq 10 $

So $ y \in [10, 15) $

Say $ y = 14 $: time = 14 < 15, distance = 840 → OK

So maximum possible driving time for friend: up to almost 15 hours, but must be greater than yours

So answer to: "How many hours can you and your friend drive per day?"

We need to report possible values or range.

But likely they want the maximum total hours under constraints.

But total time $ x + y < 15 $, so less than 15 hours

Also, your friend drives longer than you, so $ y > x $

So maximum total time approaches 15 hours, but not reaching it.

For example, $ x = 7 $, $ y = 7.9 $ → total = 14.9 < 15, $ y > x $, distance = $ 70(7) + 60(7.9) = 490 + 474 = 964 $ ≥ 600 → OK

So maximum total driving time is just under 15 hours

But the question asks: "How many hours can you and your friend drive per day?"

Since it's a system, we can say:

> You and your friend can drive up to just under 15 hours per day, with the friend driving more than you, and covering at least 600 miles.

But if they want specific numbers, perhaps find maximum possible values.

Alternatively, maybe they want how many hours each can drive, but since it's open-ended, we describe the solution set.

But let’s re-read: "How many hours can you and your friend drive per day?"

It’s ambiguous — but likely they want the maximum possible total hours under constraints.

But total must be less than 15, so maximum total is approaching 15 hours.

But to give a concrete answer, perhaps list one possible combination.

But better to summarize.

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✔ Final Answer for Problem 2:

Inequalities:
1. $ x + y < 15 $
2. $ 70x + 60y \geq 600 $
3. $ y > x $

Graph:
- Shade below $ x + y = 15 $
- Shade above $ 70x + 60y = 600 $
- Shade above $ y = x $

Feasible region exists for $ x $ from about 0 to 7, $ y $ from about 6.5 to 14.9

Maximum total driving time: less than 15 hours

Example: $ x = 7 $, $ y = 7.5 $ → total = 14.5 hours, distance = $ 70(7) + 60(7.5) = 490 + 450 = 940 $ miles

So you and your friend can drive up to just under 15 hours per day, depending on speeds and constraints.

But to answer: How many hours can you and your friend drive per day?

👉 The maximum total driving time is less than 15 hours, so they can drive up to 14.99... hours (but not 15).

But in practice, they can drive any combination where total time is less than 15 hours, friend drives longer, and total distance ≥ 600 miles.

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✔ Summary:

#### Problem 1:
- a) $ 7x + 4y < 40 $
- b) $ y \geq 5 $
- c) Graph both, feasible region bounded by those lines.
- Max wings: 2 packages
- Max hot dogs: 9 pounds

#### Problem 2:
- Inequalities:
- $ x + y < 15 $
- $ 70x + 60y \geq 600 $
- $ y > x $
- Graph all three.
- You and your friend can drive up to less than 15 hours per day, with your friend driving longer than you, and covering at least 600 miles.

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Would you like me to generate a sketch of the graphs? (I can describe how to draw them.)