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.