General Inequalities and Their Applications

1. Solve for \(r\).
Reduce any fractions to lowest terms. Don't round your answer, and don't use mixed fractions.

\(35 r-21 < -35 r+19\)

2. Solve for \(a\).
Reduce any fractions to lowest terms. Don't round your answer, and don't use mixed fractions.

\(60 a+64 \geq 80 a-92\)

3. Solve for \(t\).
Reduce any fractions to lowest terms. Don't round your answer, and don't use mixed fractions.

\(-48 t+2 \leq-71 t+14\)

4. Solve for \(w\).
Reduce any fractions to lowest terms. Don't round your answer, and don't use mixed fractions.

\(53 w+13 < 56 w+16\)

1. \(r < \frac{4}{7}\)

In conclusion, the answer is \(r < \frac{4}{7}\).

2. \(a \leq \frac{39}{5}\)

Why did the inequality sign flip when we multiplied by \(-1\)?

The inequality sign flips because we order negative numbers differently from positive numbers.

For example, \(2 < 3\). However, when we multiply both sides of the inequality by \(-1\), we see that the inequality flips, because \(-2 < -3\).

In general, if \(u < k\), then it follows that \(-u < -k\).

In conclusion, the answer is \(a \leq \frac{39}{5}\).

3. \(t \leq \frac{12}{23}\)

In conclusion, the answer is \(t \leq \frac{12}{23}\).

4. \(w > -1\)

Why did the inequality sign flip when we multiplied by -1?

The inequality sign flips because we order negative numbers differently from positive numbers.

For example, \(2 < 3\). However, when we multiply both sides of the inequality by \(-1\), we see that the inequality flips, because \(-2 < -3\).

In general, if \(u < k\), then it follows that \(-u > -k\).

In conclusion, the answer is \(w > -1\).

1. Jacque needs to buy some pizzas for a party at her office. She's ordering from a restaurant that charges a \($7.50\) delivery fee and \($14\) per pizza. She wants to buy as many pizzas as she can, and she also needs to keep the delivery fee plus the cost of the pizzas under \($60\).

Each pizza is cut into \(8\) slices, and she wonders how many total slices she can afford.

Let \(P\) represent the number of pizzas that Jacque buys.

1) Which inequality describes this scenario?

Choose 1 answer:

A. \(7.50+14 P < 60\)

B. \(7.50+14 P > 60\)

C. \(14+7.50 P < 60\)

D. \(14+7.50 P > 60\)

2) What is the largest number of slices that Jacque can afford?

2. Sofia ordered sushi for a company meeting. They change plans and increase how many people will be at the meeting, so they need at least \(100\) pieces of sushi in total.

Sofia had already ordered and paid for \(24\) pieces of sushi, so she needs to order additional sushi. The sushi comes in rolls, and each roll contains \(12\) pieces and costs \($8\).

Let \(R\) represent the number of additional rolls that Sofia orders.

1) Which inequality describes this scenario?

Choose 1 answer:

A. \(12+24 R \leq 100\)

B. \(12+24 R \geq 100\)

C. \(24+12 R \leq 100\)

D. \(24+12 R \geq 100\)

2) What is the least amount of additional money Sofia can spend to get the sushi they need?

3. Sergei runs a bakery. He needs at least \(175\) kilograms of flour in total to complete the holiday orders he's received. He only has \(34\) kilograms of flour, so he needs to buy more.

The flour he likes comes in bags that each contain \(23\) kilograms of flour. He wants to buy the smallest number of bags as possible and get the amount of flour he needs.

Let \(F\) represent the number of bags of flour that Sergei buys.

1) Which inequality describes this scenario?

A. \(34+23 F \leq 175\)

B. \(34+23 F \geq 175\)

C. \( 23+34 F \leq 175\)

D. \(23+34 F \geq 175\)

2) What is the smallest number of bags that Sergei can buy to get the amount of flour he needs?

4. The price of a train ticket consists of an initial fee of \($5\) plus a fee of \($2.75\) per stop. Julia has \($21\) and would like to travel \(50\) kilometers. She wants to know the largest number of stops she can afford to buy on a ticket.

Let \(S\) represent the number of stops that Julia buys.

1) Which inequality describes this scenario?

Choose 1 answer:

A. \(5+2.75 \cdot S \leq 21\)

B. \(5+2.75 \cdot S \geq 21\)

C. \(5+2.75 \cdot S \leq 50\)

D. \(5+2.75 \cdot S \geq 50\)

2) What is the largest number of stops that Julia can afford?

1. A. \(7.50+14 P < 60\), \(24\) Slices

Strategy

Jacque wants the delivery fee plus the cost of the pizzas to be under \($60\). We can represent this with an inequality whose structure looks something like this:

\(\text { (delivery fee })+\text { (cost of pizzas) }[ < \text { or } > ] 60\)

Then, we can solve the inequality for \(P\) to find how many pizzas Jacque can afford.

1) Which inequality?

• The delivery fee is \($7.50\).
• Each pizza costs \($14\), and \(P\) represents the number of pizzas Jacque buys, so the cost of pizzas is \(14 \cdot P\).
• Jacque wants the delivery fee plus the cost of the pizzas to be under \($60\), so the total must be less than \($60\).

\(\text { (delivery fee })+(\text { cost of pizzas })[ < \text { or } > ] 60\)

\( 7.50+14 P < 60\)

2) How many pizzas can Jacque afford?

Let's solve our inequality for \(S\)

\(7.50+14 P < 60\) Subtract \(7.50\)

\(14 P < 52.50\) Divide by \(14\)

\(P < 3.75\)

Since she can't buy partial pizzas, Jacque can afford at most \(3\) pizzas. And each pizza has \(8\) slices, so buying \(3\) pizzas gets her \(3 \cdot 8=24\) slices.

Let's check our solution

Answers

1) The inequality that describes this scenario is \(7.50+14 P < 60\).

2) Jacque can afford at most \(24\) slices.

2. D. \(24+12 R \geq 100\), \($56\) dollars

Strategy

Sofia needs the sushi she's already ordered plus the additional sushi to be at least \(100\) pieces. We can represent this with an inequality whose structure looks something like this:

\(\text { (sushi already ordered })+\text { (additional sushi) }[\leq \text { or } \geq] 100\)

Then, we can solve the inequality for \(R\) to find how many additional rolls Sofia needs to order.

1) Which inequality?

• Sofia has already ordered and paid for \(24\) pieces.
• Each roll has \(12\) pieces, and \(R\) represents the number of additional rolls, so the number of additional pieces from these rolls is \(12R\).
• The number of pieces she's already ordered plus the additional pieces needs to be greater than or equal to \(100\) pieces.

\(\text { (sushi already ordered) }+\text { (additional sushi) }[\leq \text { or } \geq] 100\)

\(24+12 R \geq 100\)

2) How many additional rolls does Sofia need?

Let's solve our inequality for \(R\):

\(24+12 R \geq 100\) Subtract \(24\)

\(12 R \geq 76\) Divide by \(12\)

\(R \geq 6 .3\)

Since she can't order partial rolls, Sofia needs to reserve \(7\) additional rolls. And each roll costs \($8\), so ordering \(7\) additional rolls costs \(7 \cdot \$ 8=\$ 56\).

Let's check our solution

Answers

1) The inequality that describes this scenario is \(24+12 R \geq 100\).

2) Sofia needs to spend \($56\) on additional sushi.

3. B. \(34+23 F \geq 175\), \(7\) bags

The flour Sergei already has plus the flour he buys must be greater than or equal to \(175\) kilograms. We can represent this with an inequality whose structure looks something like this:

\(\text { (amount he has) }+\text { (amount he buys) }[\leq \text { or } \geq] 175\)

Then, we can solve the inequality for \(F\) to find how many bags of flour Sergei needs to buy.

1) Which inequality?

• Sergei already has \(34\) kilograms of flour.
• Each bag of flour contains \(23\) kilograms, and \(F\) represents the number of bags he buys, so the amount of flour he buys is \(23 \cdot F\).
• The amount of flour he has combined with the amount of flour he buys must be greater than or equal to \(175\) kilograms.

\(\text { (amount he has) }+(\text { amount he buys) }[\leq \text { or } \geq] 175\)

\(34+23 F \geq 175\)

2) How many bags does Sergei need?

Let's solve our inequality for \(F\):

\(34+23 F \geq 175\) Subtract \(34\)

\(23 F \geq 141\) Divide by \(23\)

\(F \geq 6.13 \)

Since he can't buy a partial bag of flour, Sergei needs to buy \(7\) bags.

Let's check our solution

Answers

1) The inequality that describes this scenario is \(34+23 F \geq 175\)

2) Sergei needs to buy \(7\) bags to get the amount of flour he needs.

4. \(5+2.75 \cdot S \leq 21\), \(5\) stops

Strategy

The money Julia spends on her ticket must be less than or equal to the \($21\) she has. We can represent this with an inequality whose structure looks something like this:

\(\text { (initial fee) }+\text { (total fees for stops) }[\leq \text { or } \geq] 21\)

Then, we can solve the inequality for \(S\) to find how many stops Julia can afford.

1) Which inequality?

• The initial fee is \($5\).
• Each stop costs \($2.75\) and \(S\) represents the number of stops Julia buys, so she's spending \(2.75 \cdot S\) on stops.
• The combined amount of money she spends on her ticket must be less than or equal \($21\).

\(\text { (initial fee) }+\text { (total fees for stops) }[\leq \text { or } \geq] 21\)

\(5+2.75 \cdot S \leq 21\)

2) How many stops can Julia afford?

Let's solve our inequality for \(S\).

\(5+2.75 \cdot S \leq 21\) Subtract \(5\).

\(2.75 \cdot S \leq 16\) Divide by \(2.75\)

\(S \leq 5 .81 \)

Since she can't buy partial stops, Julia can afford at most \(5\) stops.

Let's check our solution

Answers

1) The inequality that describes this scenario is \(5+2.75 \cdot S \leq 21\)

2) Julia can afford at most \(5\) stops.