SOLVING LINEAR INEQUALITIES WORD PROBLEMS IN TWO VARIABLES

Problem 1 :

Katie has $50 in a savings account at the beginning of the summer. She wants to have at least $20 in the account by the end of the summer. She withdraws $2 each week for food, clothes, and movie tickets. Write an inequality that expresses Katie’s situation and display it on the graph below. For how many weeks can Katie withdraw money?

Solution :

Let x be the number of weeks 

50 - 2x ≥ 20

2x ≤ 30

x ≤ 15 weeks

Problem 2 :

Skate Land charges a $50 flat fee for a birthday party rental and $4 for each person. Joann has no more than $100 to budget for her party. Write an inequality that models her situation and display it on the graph below. How many people can attend Joann's party.

Solution :

Assume x people can attend the party.

y = 50 + 4x

y ≤ 100

50 + 4x ≤ 100

4x ≤ 50

x ≤ 12.5

So, 12 people can attend Joan’s party.

Problem 3 :

Sarah is selling bracelets and earrings to make money for summer vacation. The bracelets cost $2 and the earrings cost $3. She needs to make at least $60. Sarah knows she will sell more than 10 bracelets. Write inequalities to represent the income from jewelry sold and number of bracelets sold. Find two possible solutions.

Solution :

Let x be the number of bracelets sold.

Let y be the number of earrings sold.

2x + 3y ≥ 60

x > 10

If x = 11, then 2(11) + 3y ≥ 60

3y ≥ 60 - 22

3y ≥ 38

Since y is the number of earrings, x = 11 is not possible.

Problem 4 :

An online store sells digital cameras and cell phones. The store makes a $100 profit on the sale of each digital camera x and a $50 profit on the sale of each cell phone y. The store wants to make a profit of at least $300 from its sales of digital cameras and cell phones. Write and graph an inequality that represents how many digital cameras and cell phones they must sell. Identify and interpret two solutions of the inequality.

Solution :

100x + 50y  ≥ 300

50y ≥ - 100x + 300

y ≥ (-100x/50) + (300/50)

y ≥ -2x + 6

Let y = -2x + 6

(3, 0) and (0, 6)

Choosing the point above the line, applying (6, 3) we get

y ≥ -2x + 6

3 ≥ -2(6) + 6

3 ≥ -12 + 6

3 ≥ -6

Since it is true, the solution region must be above the line.

choosing two points from the solution region, we get (4, 4) and (6, 6).

100x + 50y  ≥ 300

Applying the point (4, 4), we get

100(4) + 50(4) ≥ 300

400 + 200 ≥ 300

600 ≥ 300

To have the profit more than $300 the store have to sale 4 cameras and 4 cell phones

Applying the point (6, 6), we get

100(6) + 50(6) ≥ 300

600 + 300 ≥ 300

900 ≥ 300

To have the profit more than $300 the store have to sale 6 cameras and 6 cell phones

Problem 5 :

The Reel Good Cinema is conducting a mathematical study. In its theater, there are 200 seats. Adult tickets cost $12.50 and child tickets cost $6.25. The cinema's goal is to sell at least $1500 worth of tickets for the theater. Write a system of linear inequalities that can be used to find the possible combinations of adult tickets, x, and child tickets, y, that would satisfy the cinema's goal. Graph the solution to this system of inequalities on the set of axes below. Label the solution with an S. Marta claims that selling 30 adult tickets and 80 child tickets will result in meeting the cinema's goal. Explain whether she is correct or incorrect, based on the graph drawn.

Solution :

Total number of seats = 200

From this, we understand that total number of seats cannot be more than 200.

x + y ≤ 200

12.50x + 6.25y ≥ 1500

Finding x and y intercepts of x + y ≤ 200 :

y = -x + 200

Choosing one of the points above the line which is (200, 200), we get

200 + 200 ≤ 200

400 ≤ 200

Since it is false, we have to shade the region which is below the line.

Finding x and y intercepts of 12.50x + 6.25y ≥ 1500 :

Choosing one of the points above the line which is (120, 240), we get

12.50(120) + 6.25(240) ≥ 1500

1500 + 1500 ≥ 1500

3000 ≥ 1500

True

Since it is true, we have to shade the region which is above the line.

12.50x + 6.25y ≥ 1500

Adult tickets = 30

Child tickets = 80

12.50(30) + 6.25(80) ≥ 1500

375 + 500 ≥ 1500

875 ≥ 1500

False

So, it is not a solution.