The document provides examples for solving systems of linear equations by substitution. It explains the substitution method and walks through three examples step-by-step. The key steps are: 1) solve one equation for one variable, 2) substitute the expression into the other equation, 3) solve the resulting equation for the second variable, 4) substitute values back into the original equations to check. An example application compares the total costs of two cell phone plans over time.
Polynomials And Linear Equation of Two VariablesAnkur Patel
A complete description of polynomials and also various methods to solve the Linear equation of two variables by substitution, cross multiplication and elimination methods.
For polynomials it also contains the description of monomials, binomials etc.
If you are looking for math video tutorials (with voice recording), you may download it on our YouTube Channel. Don't forget to SUBSCRIBE for you to get updated on our upcoming videos.
https://tinyurl.com/y9muob6q
Also, please do visit our page, LIKE and FOLLOW us on Facebook!
https://tinyurl.com/ycjp8r7u
https://tinyurl.com/ybo27k2u
Social Networking Presentation For Chamber Of CommerceBARBARA ROZGONYI
Social Networking: Connecting Business, Life and Branding for Chamber of Commerce groups.
Developed and presented by Barbara Rozgonyi, this 30-minute presentation introduces groups to the world of social media via LinkedIn, Facebook and Twitter. This presentation drew one of the best attended luncheons of the year. For more information, contact Barbara Rozgonyi at 630.207.7530 or @wiredprworks on twitter.com. Barbara publishes http://wiredPRworks.com and founded Social Media Club Chicago.
Polynomials And Linear Equation of Two VariablesAnkur Patel
A complete description of polynomials and also various methods to solve the Linear equation of two variables by substitution, cross multiplication and elimination methods.
For polynomials it also contains the description of monomials, binomials etc.
If you are looking for math video tutorials (with voice recording), you may download it on our YouTube Channel. Don't forget to SUBSCRIBE for you to get updated on our upcoming videos.
https://tinyurl.com/y9muob6q
Also, please do visit our page, LIKE and FOLLOW us on Facebook!
https://tinyurl.com/ycjp8r7u
https://tinyurl.com/ybo27k2u
Social Networking Presentation For Chamber Of CommerceBARBARA ROZGONYI
Social Networking: Connecting Business, Life and Branding for Chamber of Commerce groups.
Developed and presented by Barbara Rozgonyi, this 30-minute presentation introduces groups to the world of social media via LinkedIn, Facebook and Twitter. This presentation drew one of the best attended luncheons of the year. For more information, contact Barbara Rozgonyi at 630.207.7530 or @wiredprworks on twitter.com. Barbara publishes http://wiredPRworks.com and founded Social Media Club Chicago.
Building an Enterprise Content Management solution on top of liferayAndrea Di Giorgi
Documents, data tables, wiki, message boards and so on: as we all know, Liferay provides a series of native portlets for storing and managing several types of content in your organization. But sometimes more advanced features are required, and the powerful frameworks that lay under the hood can be leveraged in order to reach your custom needs. This session presents SMC's first steps in building a Liferay-based Enterprise Content Management solution, which introduces a whole new set of functionalities for documents and other types of assets. But the potential is endless, and plans are to add even more features and have a complete ECM solution built on top of the Liferay platform.
This presentation was originally done for CTO School, a meetup for CTOs, VP Eng, or Tech Leads that are interested in improving their skills and learning from each other.
It focuses on how to be helpful to others and bringing a targeted framework to your networking.
Additional resources from this talk can be found at http://life-longlearner.com/CTOSchool
Moved to https://slidr.io/azzazzel/leveraging-osgi-to-create-extensible-plugi...Milen Dyankov
This slide deck will be removed from here in the future. It has been moved to : https://slidr.io/azzazzel/leveraging-osgi-to-create-extensible-plugins-for-liferay-6-2
EclipseCon Europe 2015 - liferay modularity patterns using OSGi -Rafik HarabiRafik HARABI
This presentation focus on modern architecting and development patterns with examples.
Liferay 7 come with new modular architecture based on OSGi framework. This new architecture will change the way of using and extending Liferay: It provides flexible options to customize Liferay portal and build applications on the top of it.
After introducing the new modular architecture and the Liferay module framework, the presentation will focus on the modern patterns of bundles development, portal customization patterns and integration with third parties using the power of the OSGi framework.
Corresponding article on http://blog.jayare.eu/articles/social-media-marketing.
A presentation about the possibilities of social media, how to use it to generate traffic and increase your online visibility.
You can find the corresponding article on http://blog.jayare.eu/articles/social-media-marketing
See what pitfalls companies are facing when running Liferay portal. In the previous year, our company has audited 5 real-life projects based on Liferay Portal which are now running in production mode and serving many users. The audits were focused on architecture, infrastructure, technical design and implementation. During the presentation, we will show you common anti-patterns we have found during the audits and their impacts and consequences on the portal.
Presentatie tijdens de iProfs-Liefray Roadshow 3 October 2013 in Solt Zeist: Hoogtepunten & samenvatting van het Liferay Portal Solutions Forum van 24 september 2013 in Frankfurt
This learner's module discusses and help the students about the topic Systems of Linear Inequalities. It includes definition, examples, applications of Systems of Linear Inequalities.
3. Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.
4. Step 1 Solve for one variable in at least one equation, if necessary. Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute that value into one of the original equations and solve. Write the values from steps 3 and 4 as an ordered pair, ( x , y ), and check. Step 5 Step 4 Step 3 Step 2 Solving Systems of Equations by Substitution
5. Solve the system by substitution. Example 1A: Solving a System of Linear Equations by Substitution y = 3 x y = x – 2 Step 1 y = 3 x y = x – 2 Both equations are solved for y. Substitute 3x for y in the second equation. Solve for x. Subtract x from both sides and then divide by 2. Step 2 y = x – 2 3 x = x – 2 Step 3 – x – x 2 x = –2 2 x = –2 2 2 x = –1
6. Solve the system by substitution. Example 1A Continued Write one of the original equations. Substitute –1 for x. Check Substitute ( – 1, – 3) into both equations in the system. Write the solution as an ordered pair. Step 4 y = 3 x y = 3 ( – 1) y = –3 Step 5 ( – 1, –3 ) y = 3 x – 3 3 (–1) – 3 –3 y = x – 2 – 3 –1 – 2 – 3 –3
7. Solve the system by substitution. Example 1B: Solving a System of Linear Equations by Substitution y = x + 1 4 x + y = 6 Step 1 y = x + 1 The first equation is solved for y. Substitute x + 1 for y in the second equation. Subtract 1 from both sides. 5 x + 1 = 6 Simplify. Solve for x. Divide both sides by 5. Step 2 4 x + y = 6 4 x + ( x + 1) = 6 Step 3 – 1 –1 5 x = 5 5 5 x = 1 5 x = 5
8. Solve the system by substitution. Example1B Continued Write one of the original equations. Substitute 1 for x. Check Substitute (1, 2) into both equations in the system. Write the solution as an ordered pair. Step 4 y = x + 1 y = 1 + 1 y = 2 Step 5 ( 1, 2 ) y = x + 1 2 1 + 1 2 2 4 x + y = 6 4 (1) + 2 6 6 6
9. Solve the system by substitution. Example 1C: Solving a System of Linear Equations by Substitution x + 2 y = –1 x – y = 5 Step 1 x + 2 y = – 1 Solve the first equation for x by subtracting 2y from both sides. Substitute – 2 y – 1 for x in the second equation. – 3 y – 1 = 5 Simplify. Step 2 x – y = 5 (–2 y – 1) – y = 5 − 2 y −2 y x = – 2 y – 1
10. Example 1C Continued Step 3 – 3 y – 1 = 5 Add 1 to both sides. Solve for y. Divide both sides by –3. Step 5 ( 3 , –2 ) Write one of the original equations. Substitute – 2 for y. Subtract 2 from both sides. Write the solution as an ordered pair. +1 +1 – 3 y = 6 – 3 y = 6 – 3 –3 y = –2 Step 4 x – y = 5 x – (–2) = 5 x + 2 = 5 – 2 –2 x = 3
11. Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.
12. When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. Caution
13. Example 2: Using the Distributive Property y + 6 x = 11 3 x + 2 y = –5 Solve by substitution. Solve the first equation for y by subtracting 6x from each side. Substitute – 6x + 11 for y in the second equation. Distribute 2 to the expression in parenthesis. 3 x + 2 (–6 x + 11) = –5 Step 1 y + 6 x = 11 – 6 x – 6 x y = –6 x + 11 3 x + 2 (–6 x + 11) = –5 3 x + 2 y = –5 Step 2
14. Step 3 Example 2 Continued y + 6 x = 11 3 x + 2 y = –5 Solve by substitution. 3 x + 2 (–6 x ) + 2 (11) = –5 – 9 x + 22 = –5 Simplify. Solve for x. Subtract 22 from both sides. Divide both sides by –9. x = 3 3 x – 12 x + 22 = –5 – 9 x = –27 – 22 –22 – 9 x = –27 – 9 –9
15. Substitute 3 for x. y + 6 (3) = 11 Subtract 18 from each side. y + 18 = 11 Step 5 ( 3 , –7 ) Write the solution as an ordered pair. Simplify. Example 2 Continued y + 6 x = 11 3 x + 2 y = –5 Solve by substitution. Write one of the original equations. Step 4 y + 6 x = 11 – 18 – 18 y = –7
16. Example 2: Consumer Economics Application Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Write an equation for each option. Let t represent the total amount paid and m represent the number of months.
17. Example 2 Continued Total paid is sign-up fee plus payment amount for each month. Both equations are solved for t. Substitute 50 + 20m for t in the second equation. Option 1 t = $50 + $20 m Option 2 t = $30 + $25 m Step 1 t = 50 + 20 m t = 30 + 25 m Step 2 50 + 20 m = 30 + 25 m
18. Solve for m. Subtract 20m from both sides. 50 = 30 + 5 m Subtract 30 from both sides. 20 = 5 m Divide both sides by 5. Write one of the original equations. t = 30 + 25 (4) Substitute 4 for m. Simplify. Example 2 Continued Step 3 50 + 20 m = 30 + 25 m – 20 m – 20 m – 30 –30 Step 4 t = 30 + 25 m t = 30 + 100 t = 130 5 5 m = 4 20 = 5 m
19. Step 5 ( 4 , 130 ) Write the solution as an ordered pair. In 4 months, the total cost for each option would be the same $130. Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330. Example 2 Continued Option 1: t = 50 + 20(12) = 290 Option 2: t = 30 + 25(12) = 330 If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.