graphing systems homework

4.7 Graphing Systems of Linear Inequalities

Learning objectives.

By the end of this section, you will be able to:

Determine whether an ordered pair is a solution of a system of linear inequalities

• Solve a system of linear inequalities by graphing
• Solve applications of systems of inequalities

Be Prepared 4.19

Before you get started, take this readiness quiz.

Solve the inequality 2 a < 5 a + 12 . 2 a < 5 a + 12 . If you missed this problem, review Example 2.52 .

Be Prepared 4.20

Determine whether the ordered pair ( 3 , 1 2 ) ( 3 , 1 2 ) is a solution to the system y > 2 x + 3 . y > 2 x + 3 . If you missed this problem, review Example 3.34 .

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

System of Linear Inequalities

Two or more linear inequalities grouped together form a system of linear inequalities.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) ( x , y ) that make both inequalities true.

Solutions of a System of Linear Inequalities

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true.

The solution of a system of linear inequalities is shown as a shaded region in the x, y coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example 4.53

Determine whether the ordered pair is a solution to the system { x + 4 y ≥ 10 3 x − 2 y < 12 . { x + 4 y ≥ 10 3 x − 2 y < 12 .

ⓐ ( −2 , 4 ) ( −2 , 4 ) ⓑ ( 3 , 1 ) ( 3 , 1 )

ⓐ Is the ordered pair ( −2 , 4 ) ( −2 , 4 ) a solution?

The ordered pair ( −2 , 4 ) ( −2 , 4 ) made both inequalities true. Therefore ( −2 , 4 ) ( −2 , 4 ) is a solution to this system.

ⓑ Is the ordered pair ( 3 , 1 ) ( 3 , 1 ) a solution?

The ordered pair ( 3 , 1 ) ( 3 , 1 ) made one inequality true, but the other one false. Therefore ( 3 , 1 ) ( 3 , 1 ) is not a solution to this system.

Try It 4.105

Determine whether the ordered pair is a solution to the system: { x − 5 y > 10 2 x + 3 y > −2 . { x − 5 y > 10 2 x + 3 y > −2 .

ⓐ ( 3 , −1 ) ( 3 , −1 ) ⓑ ( 6 , −3 ) ( 6 , −3 )

Try It 4.106

Determine whether the ordered pair is a solution to the system: { y > 4 x − 2 4 x − y < 20 . { y > 4 x − 2 4 x − y < 20 .

ⓐ ( −2 , 1 ) ( −2 , 1 ) ⓑ ( 4 , −1 ) ( 4 , −1 )

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

Example 4.54

How to solve a system of linear inequalities by graphing.

Solve the system by graphing: { y ≥ 2 x − 1 y < x + 1 . { y ≥ 2 x − 1 y < x + 1 .

Try It 4.107

Solve the system by graphing: { y < 3 x + 2 y > − x − 1 . { y < 3 x + 2 y > − x − 1 .

Try It 4.108

Solve the system by graphing: { y < − 1 2 x + 3 y < 3 x − 4 . { y < − 1 2 x + 3 y < 3 x − 4 .

Solve a system of linear inequalities by graphing.

• Graph the boundary line.
• Shade in the side of the boundary line where the inequality is true.
• Shade in the side of that boundary line where the inequality is true.
• Step 3. The solution is the region where the shading overlaps.
• Step 4. Check by choosing a test point.

Example 4.55

Solve the system by graphing: { x − y > 3 y < − 1 5 x + 4 . { x − y > 3 y < − 1 5 x + 4 .

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Try It 4.109

Solve the system by graphing: { x + y ≤ 2 y ≥ 2 3 x − 1 . { x + y ≤ 2 y ≥ 2 3 x − 1 .

Try It 4.110

Solve the system by graphing: { 3 x − 2 y ≤ 6 y > − 1 4 x + 5 . { 3 x − 2 y ≤ 6 y > − 1 4 x + 5 .

Example 4.56

Solve the system by graphing: { x − 2 y < 5 y > −4 . { x − 2 y < 5 y > −4 .

The point ( 0 , 0 ) ( 0 , 0 ) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice—which appears as the darkest shaded region.

Try It 4.111

Solve the system by graphing: { y ≥ 3 x − 2 y < − 1 . { y ≥ 3 x − 2 y < − 1 .

Try It 4.112

Solve the system by graphing: { x > −4 x − 2 y ≥ −4 . { x > −4 x − 2 y ≥ −4 .

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in the next example.

Example 4.57

Solve the system by graphing: { 4 x + 3 y ≥ 12 y < − 4 3 x + 1 . { 4 x + 3 y ≥ 12 y < − 4 3 x + 1 .

There is no point in both shaded regions, so the system has no solution.

Try It 4.113

Solve the system by graphing: { 3 x − 2 y ≥ 12 y ≥ 3 2 x + 1 . { 3 x − 2 y ≥ 12 y ≥ 3 2 x + 1 .

Try It 4.114

Solve the system by graphing: { x + 3 y > 8 y < − 1 3 x − 2 . { x + 3 y > 8 y < − 1 3 x − 2 .

Some systems of linear inequalities where the boundary lines are parallel will have a solution. We’ll see this in the next example.

Example 4.58

Solve the system by graphing: { y > 1 2 x − 4 x − 2 y < −4 . { y > 1 2 x − 4 x − 2 y < −4 .

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice which is also the solution to x − 2 y < −4 . x − 2 y < −4 .

Try It 4.115

Solve the system by graphing: { y ≥ 3 x + 1 −3 x + y ≥ −4 . { y ≥ 3 x + 1 −3 x + y ≥ −4 .

Try It 4.116

Solve the system by graphing: { y ≤ − 1 4 x + 2 x + 4 y ≤ 4 . { y ≤ − 1 4 x + 2 x + 4 y ≤ 4 .

Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so we add inequalities to the system as additional requirements.

Example 4.59

Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her $4 and each large photo costs her $10. She doesn’t want to spend more than $200 on photos to display.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could she display 10 small and 20 large photos? ⓓ Could she display 20 small and 10 large photos?

ⓑ Since x ≥ 0 x ≥ 0 and y ≥ 0 y ≥ 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly-shaded section are included in the solution as are the points on the x -axis from (25, 0) to (55, 0).

ⓒ To determine if 10 small and 20 large photos would work, we look at the graph to see if the point (10, 20) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is not, Christy would not display 10 small and 20 large photos.

ⓓ To determine if 20 small and 10 large photos would work, we look at the graph to see if the point (20, 10) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is, so Christy could choose to display 20 small and 10 large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

Try It 4.117

A trailer can carry a maximum weight of 160 pounds and a maximum volume of 15 cubic feet. A microwave oven weighs 30 pounds and has 2 cubic feet of volume, while a printer weighs 20 pounds and has 3 cubic feet of space.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could 4 microwaves and 2 printers be carried on this trailer? ⓓ Could 7 microwaves and 3 printers be carried on this trailer?

Try It 4.118

Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost $2 and the answer sheets cost $1. Mary’s budget for these supplies allows for a maximum cost of $400.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could Mary purchase 100 pencils and 100 answer sheets? ⓓ Could Mary purchase 150 pencils and 150 answer sheets?

When we use variables other than x and y to define an unknown quantity, we must change the names of the axes of the graph as well.

Example 4.60

Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs $0.50.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could he eat 3 hamburgers and 1 cookie? ⓓ Could he eat 2 hamburgers and 4 cookies?

ⓑ Since h > = 0 h > = 0 and c > = 0 c > = 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

Graph 1.40 h + 0.50 c ≤ 5 . 1.40 h + 0.50 c ≤ 5 . The boundary line is 1.40 h + 0.50 c = 5 . 1.40 h + 0.50 c = 5 . We test (0, 0) and it makes the inequality true. We shade the side of the line that includes (0, 0).

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x -axis from (5, 0) to (10, 0).

ⓒ To determine if 3 hamburgers and 1 cookie would meet Omar’s criteria, we see if the point (3, 2) is in the solution region. It is, so Omar might choose to eat 3 hamburgers and 1 cookie.

ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is, Omar might choose to eat 2 hamburgers and 4 cookies.

We could also test the possible solutions by substituting the values into each inequality.

Try It 4.119

Tenison needs to eat at least an extra 1,000 calories a day to prepare for running a marathon. He has only $25 to spend on the extra food he needs and will spend it on $0.75 donuts which have 360 calories each and $2 energy drinks which have 110 calories.

ⓐ Write a system of inequalities that models this situation. ⓑ Graph the system. ⓒ Can he buy 8 donuts and 4 energy drinks and satisfy his caloric needs? ⓓ Can he buy 1 donut and 3 energy drinks and satisfy his caloric needs?

Try It 4.120

Philip’s doctor tells him he should add at least 1,000 more calories per day to his usual diet. Philip wants to buy protein bars that cost $1.80 each and have 140 calories and juice that costs $1.25 per bottle and have 125 calories. He doesn’t want to spend more than $12.

ⓐ Write a system of inequalities that models this situation. ⓑ Graph the system. ⓒ Can he buy 3 protein bars and 5 bottles of juice? ⓓ Can he buy 5 protein bars and 3 bottles of juice?

Access these online resources for additional instruction and practice with solving systems of linear inequalities by graphing.

• Solving Systems of Linear Inequalities by Graphing
• Systems of Linear Inequalities

Section 4.7 Exercises

Practice makes perfect.

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

In the following exercises, determine whether each ordered pair is a solution to the system.

{ 3 x + y > 5 2 x − y ≤ 10 { 3 x + y > 5 2 x − y ≤ 10

ⓐ ( 3 , −3 ) ( 3 , −3 ) ⓑ ( 7 , 1 ) ( 7 , 1 )

{ 4 x − y < 10 −2 x + 2 y > −8 { 4 x − y < 10 −2 x + 2 y > −8

ⓐ ( 5 , −2 ) ( 5 , −2 ) ⓑ ( −1 , 3 ) ( −1 , 3 )

{ y > 2 3 x − 5 x + 1 2 y ≤ 4 { y > 2 3 x − 5 x + 1 2 y ≤ 4

ⓐ (6, −4) (6, −4) ⓑ (3, 0) (3, 0)

{ y < 3 2 x + 3 3 4 x − 2 y < 5 { y < 3 2 x + 3 3 4 x − 2 y < 5

ⓐ ( −4 , −1 ) ( −4 , −1 ) ⓑ (8, 3) (8, 3)

{ 7 x + 2 y > 14 5 x − y ≤ 8 { 7 x + 2 y > 14 5 x − y ≤ 8

ⓐ (2, 3) (2, 3) ⓑ (7, −1) (7, −1)

{ 6 x − 5 y < 20 −2 x + 7 y > −8 { 6 x − 5 y < 20 −2 x + 7 y > −8

ⓐ (1, −3) (1, −3) ⓑ (−4, 4) (−4, 4)

In the following exercises, solve each system by graphing.

{ y ≤ 3 x + 2 y > x − 1 { y ≤ 3 x + 2 y > x − 1

{ y < − 2 x + 2 y ≥ − x − 1 { y < − 2 x + 2 y ≥ − x − 1

{ y < 2 x − 1 y ≤ − 1 2 x + 4 { y < 2 x − 1 y ≤ − 1 2 x + 4

{ y ≥ − 2 3 x + 2 y > 2 x − 3 { y ≥ − 2 3 x + 2 y > 2 x − 3

x − y > 1 y < − 1 4 x + 3 x − y > 1 y < − 1 4 x + 3

{ x + 2 y < 4 y < x − 2 { x + 2 y < 4 y < x − 2

{ 3 x − y ≥ 6 y ≥ − 1 2 x { 3 x − y ≥ 6 y ≥ − 1 2 x

{ 2 x + 4 y ≥ 8 y ≤ 3 4 x { 2 x + 4 y ≥ 8 y ≤ 3 4 x

{ 2 x − 5 y < 10 3 x + 4 y ≥ 12 { 2 x − 5 y < 10 3 x + 4 y ≥ 12

{ 3 x − 2 y ≤ 6 −4 x − 2 y > 8 { 3 x − 2 y ≤ 6 −4 x − 2 y > 8

{ 2 x + 2 y > −4 − x + 3 y ≥ 9 { 2 x + 2 y > −4 − x + 3 y ≥ 9

{ 2 x + y > −6 − x + 2 y ≥ −4 { 2 x + y > −6 − x + 2 y ≥ −4

{ x − 2 y < 3 y ≤ 1 { x − 2 y < 3 y ≤ 1

{ x − 3 y > 4 y ≤ − 1 { x − 3 y > 4 y ≤ − 1

{ y ≥ − 1 2 x − 3 x ≤ 2 { y ≥ − 1 2 x − 3 x ≤ 2

{ y ≤ − 2 3 x + 5 x ≥ 3 { y ≤ − 2 3 x + 5 x ≥ 3

{ y ≥ 3 4 x − 2 y < 2 { y ≥ 3 4 x − 2 y < 2

{ y ≤ − 1 2 x + 3 y < 1 { y ≤ − 1 2 x + 3 y < 1

{ 3 x − 4 y < 8 x < 1 { 3 x − 4 y < 8 x < 1

{ −3 x + 5 y > 10 x > −1 { −3 x + 5 y > 10 x > −1

{ x ≥ 3 y ≤ 2 { x ≥ 3 y ≤ 2

{ x ≤ −1 y ≥ 3 { x ≤ −1 y ≥ 3

{ 2 x + 4 y > 4 y ≤ − 1 2 x − 2 { 2 x + 4 y > 4 y ≤ − 1 2 x − 2

{ x − 3 y ≥ 6 y > 1 3 x + 1 { x − 3 y ≥ 6 y > 1 3 x + 1

{ −2 x + 6 y < 0 6 y > 2 x + 4 { −2 x + 6 y < 0 6 y > 2 x + 4

{ −3 x + 6 y > 12 4 y ≤ 2 x − 4 { −3 x + 6 y > 12 4 y ≤ 2 x − 4

{ y ≥ −3 x + 2 3 x + y > 5 { y ≥ −3 x + 2 3 x + y > 5

{ y ≥ 1 2 x − 1 −2 x + 4 y ≥ 4 { y ≥ 1 2 x − 1 −2 x + 4 y ≥ 4

{ y ≤ − 1 4 x − 2 x + 4 y < 6 { y ≤ − 1 4 x − 2 x + 4 y < 6

{ y ≥ 3 x − 1 −3 x + y > −4 { y ≥ 3 x − 1 −3 x + y > −4

{ 3 y > x + 2 −2 x + 6 y > 8 { 3 y > x + 2 −2 x + 6 y > 8

{ y < 3 4 x − 2 −3 x + 4 y < 7 { y < 3 4 x − 2 −3 x + 4 y < 7

In the following exercises, translate to a system of inequalities and solve.

Caitlyn sells her drawings at the county fair. She wants to sell at least 60 drawings and has portraits and landscapes. She sells the portraits for $15 and the landscapes for $10. She needs to sell at least $800 worth of drawings in order to earn a profit.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Will she make a profit if she sells 20 portraits and 35 landscapes? ⓓ Will she make a profit if she sells 50 portraits and 20 landscapes?

Jake does not want to spend more than $50 on bags of fertilizer and peat moss for his garden. Fertilizer costs $2 a bag and peat moss costs $5 a bag. Jake’s van can hold at most 20 bags.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Can he buy 15 bags of fertilizer and 4 bags of peat moss? ⓓ Can he buy 10 bags of fertilizer and 10 bags of peat moss?

Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than $500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is $3 and for a package the cost is $7.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Can she mail 60 cards and 26 packages? ⓓ Can she mail 90 cards and 40 packages?

Juan is studying for his final exams in chemistry and algebra. he knows he only has 24 hours to study, and it will take him at least three times as long to study for algebra than chemistry.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Can he spend 4 hours on chemistry and 20 hours on algebra? ⓓ Can he spend 6 hours on chemistry and 18 hours on algebra?

Jocelyn is pregnant and so she needs to eat at least 500 more calories a day than usual. When buying groceries one day with a budget of $15 for the extra food, she buys bananas that have 90 calories each and chocolate granola bars that have 150 calories each. The bananas cost $0.35 each and the granola bars cost $2.50 each.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could she buy 5 bananas and 6 granola bars? ⓓ Could she buy 3 bananas and 4 granola bars?

Mark is attempting to build muscle mass and so he needs to eat at least an additional 80 grams of protein a day. A bottle of protein water costs $3.20 and a protein bar costs $1.75. The protein water supplies 27 grams of protein and the bar supplies 16 gram. If he has $10 dollars to spend

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could he buy 3 bottles of protein water and 1 protein bar? ⓓ Could he buy no bottles of protein water and 5 protein bars?

Jocelyn desires to increase both her protein consumption and caloric intake. She desires to have at least 35 more grams of protein each day and no more than an additional 200 calories daily. An ounce of cheddar cheese has 7 grams of protein and 110 calories. An ounce of parmesan cheese has 11 grams of protein and 22 calories.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could she eat 1 ounce of cheddar cheese and 3 ounces of parmesan cheese? ⓓ Could she eat 2 ounces of cheddar cheese and 1 ounce of parmesan cheese?

Mark is increasing his exercise routine by running and walking at least 4 miles each day. His goal is to burn a minimum of 1500 calories from this exercise. Walking burns 270 calories/mile and running burns 650 calories.

ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could he meet his goal by walking 3 miles and running 1 mile? ⓓ Could he meet his goal by walking 2 miles and running 2 miles?

Writing Exercises

Graph the inequality x − y ≥ 3 . x − y ≥ 3 . How do you know which side of the line x − y = 3 x − y = 3 should be shaded?

Graph the system { x + 2 y ≤ 6 y ≥ − 1 2 x − 4 . { x + 2 y ≤ 6 y ≥ − 1 2 x − 4 . What does the solution mean?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/4-7-graphing-systems-of-linear-inequalities

© Jul 24, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

• Quick Facts
• Sights & Attractions
• Tsarskoe Selo
• Oranienbaum
• Foreign St. Petersburg
• Restaurants & Bars
• Accommodation Guide
• St. Petersburg Hotels
• Serviced Apartments
• Bed and Breakfasts
• Private & Group Transfers
• Airport Transfers
• Concierge Service
• Russian Visa Guide
• Request Visa Support
• Walking Tours
• River Entertainment
• Public Transportation
• Travel Cards
• Essential Shopping Selection
• Business Directory
• Photo Gallery
• Video Gallery
• 360° Panoramas
• Moscow Hotels
• Moscow.Info
• Imperial Estates
• Oranienbaum (Lomonosov)

Visiting Oranienbaum and Lomonosov

Oranienbaum is perhaps the least visited of St. Petersburg's suburban palaces, despite the rich and unique treasures it has to offer. This can partly be attributed to the decades of neglect that have left most of the buildings in critical condition and allowed the park to be overrun by nature. Thankfully, the estate has now come under the auspices of Peterhof's management, and a massive renovation programme is already underway. For the next several years, however, it will be necessary to check which parts of the park and which buildings are currently open to the public.

Oranienbaum is only 12km along the coast from Peterhof, so it is possible for those with some stamina to combine the two in a single daytrip. A marshrutka minibus (K-348) connects the two estates. From St. Petersburg it is possible to take a minibus (K-300) from Avtovo Metro Station, or suburban trains from Baltiskiy Station. Either way, the journey takes a little under one hour.

Oranienbaum Park

Located on a slight incline, this pretty landscaped park is slowly being restored to its former glories, and contains a number of unusual architectural features.

Grand Menshikov Palace

This superbly elegant baroque palace was originally built for Peter the Great's closest adviser, and was the only suburban palace to survive World War 2 intact.

Chinese Palace

The favorite retreat of Catherine the Great, this small but exquisitely designed neoclassical villa has superbly restored interiors in "Oriental" style.

Palace of Peter III

This modest two-storey building was built for the future Peter III, and was the centre of Petershtadt, his peculiar model fortress of which he styled himself the Commandant.

Built as part of a Summer Palace for Peter III, this simple two-storey block by Antonio Rinaldi was later used as a Lutheran church, and is now an exhibition and concert hall.

We can help you make the right choice from hundreds of St. Petersburg hotels and hostels.

Live like a local in self-catering apartments at convenient locations in St. Petersburg.

Comprehensive solutions for those who relocate to St. Petersburg to live, work or study.

Maximize your time in St. Petersburg with tours expertly tailored to your interests.

Get around in comfort with a chauffeured car or van to suit your budget and requirements.

Book a comfortable, well-maintained bus or a van with professional driver for your group.

Navigate St. Petersburg’s dining scene and find restaurants to remember.

Need tickets for the Mariinsky, the Hermitage, a football game or any event? We can help.

Get our help and advice choosing services and options to plan a prefect train journey.

Let our meeting and events experts help you organize a superb event in St. Petersburg.

We can find you a suitable interpreter for your negotiations, research or other needs.

Get translations for all purposes from recommended professional translators.

Accurate Sampling-Based Cardinality Estimation for Complex Graph Queries

New citation alert added.

This alert has been successfully added and will be sent to:

You will be notified whenever a record that you have chosen has been cited.

To manage your alert preferences, click on the button below.

New Citation Alert!

Please log in to your account

Information & Contributors

Bibliometrics & citations, view options, index terms.

Information systems

Data management systems

Database management system engines

Database query processing

Recommendations

Weighted distinct sampling: cardinality estimation for spj queries.

SPJ (select-project-join) queries form the backbone of many SQL queries used in practice. Accurate cardinality estimation of these queries is thus an important problem, with applications in query optimization, approximate query processing, and data ...

Equivalence and minimization of conjunctive queries under combined semantics

The problems of query containment, equivalence, and minimization are fundamental problems in the context of query processing and optimization. In their classic work [2] published in 1977, Chandra and Merlin solved the three problems for the language of ...

Synopses for query optimization: A space-complexity perspective

Database systems use precomputed synopses of data to estimate the cost of alternative plans during query optimization. A number of alternative synopsis structures have been proposed, but histograms are by far the most commonly used. While histograms ...

Information

Published in.

Association for Computing Machinery

New York, NY, United States

Publication History

Check for updates, author tags.

• cardinality estimation
• conjunctive queries
• query planning
• Research-article

Contributors

Other metrics, bibliometrics, article metrics.

• 0 Total Citations
• 31 Total Downloads
• Downloads (Last 12 months) 31
• Downloads (Last 6 weeks) 31

View options

View or Download as a PDF file.

View online with eReader .

Login options

Check if you have access through your login credentials or your institution to get full access on this article.

Full Access

Share this publication link.

Copying failed.

Share on social media

Affiliations, export citations.

• Please download or close your previous search result export first before starting a new bulk export. Preview is not available. By clicking download, a status dialog will open to start the export process. The process may take a few minutes but once it finishes a file will be downloadable from your browser. You may continue to browse the DL while the export process is in progress. Download
• Download citation
• Copy citation

We are preparing your search results for download ...

We will inform you here when the file is ready.

Your file of search results citations is now ready.

Your search export query has expired. Please try again.