This document provides examples of solving linear equations with one variable. It begins with examples of solving equations with the variable on both sides, including subtracting like terms to isolate the variable. It then gives examples of solving more complex multi-step equations, showing the process of combining like terms and using properties of equality. Finally, it provides a word problem example involving costs for flower bouquets with different variables representing fees and costs per item.
Mathematics 9 Lesson 1-A: Solving Quadratic Equations by Completing the SquareJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson Solving Quadratic Equations using Completing the Square. It also discusses the steps in solving quadratic equations using the method of Completing the Square.
Mathematics 9 Lesson 1-A: Solving Quadratic Equations by Completing the SquareJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson Solving Quadratic Equations using Completing the Square. It also discusses the steps in solving quadratic equations using the method of Completing the Square.
Materi Bilangan Bulat Matematika Kelas 7
Terdiri dari :
Penjumlahan Bilangan Bulat
Pengurangan Bilangan Bulat
Perkalian Bilangan Bulat
Pembagian Bilangan Bulat
Pecahan Bilangan Bulat
Desimal
KPK dan FPB
Contoh Soal Bilangan Bulat
Getting More Business from Your Members with Electronic Strategies (Credit Un...NAFCU Services Corporation
In this 2012 Strategic Growth Conference session, learn how effectively communicating with your members and potential members has recently become more complicated. Explore how you can successfully integrate electronic response communications with conventional communication vehicles to stimulate your credit union’s growth. Discover the importance of your members’ data and how segmentation can make your growth strategies more valuable. You will walk away with three tools that you can start implementing immediately to make your communication strategies work better for you while developing your members’ engagement and helping your credit union grow. More info at: www.nafcu.org/cathedral
Materi Bilangan Bulat Matematika Kelas 7
Terdiri dari :
Penjumlahan Bilangan Bulat
Pengurangan Bilangan Bulat
Perkalian Bilangan Bulat
Pembagian Bilangan Bulat
Pecahan Bilangan Bulat
Desimal
KPK dan FPB
Contoh Soal Bilangan Bulat
Getting More Business from Your Members with Electronic Strategies (Credit Un...NAFCU Services Corporation
In this 2012 Strategic Growth Conference session, learn how effectively communicating with your members and potential members has recently become more complicated. Explore how you can successfully integrate electronic response communications with conventional communication vehicles to stimulate your credit union’s growth. Discover the importance of your members’ data and how segmentation can make your growth strategies more valuable. You will walk away with three tools that you can start implementing immediately to make your communication strategies work better for you while developing your members’ engagement and helping your credit union grow. More info at: www.nafcu.org/cathedral
Are we ready to make the UK the best country to grow old in?
One year ago, the House of Lords Committee on Public Services and Demographic Change produced a hard-hitting report which argued that the Government and society was “woefully underprepared” for a rapidly ageing population.
On the first anniversary of the ‘Ready for Ageing?’ report, we are in the unenviable position that sees the United Kingdom ranked unlucky number 13 in a global index of the best countries in the world to grow old in. The principal recommendations in the ‘Ready for Ageing?’ report have not yet been properly addressed or acted on.
In his October 2013 speech on ‘The Forgotten Million’, Secretary of State for Health, Jeremy Hunt MP, set down a challenge that the UK should in fact aspire to be best country to grow old in, but the question remains: why are our public services so poorly prepared for major demographic change, and what as a society can we do to ensure future generations of older people thrive in later life?
Lord Filkin, Chair of the Committee on Public Services and Demographic Change, hosted a House of Lords breakfast debate looking forward to 2030, a date by which there will be 50% more people aged 65 and over in England and a doubling in the numbers of people aged 85 and over. As a society, we need to prepare for the next 15 years right now and certainly in the next Parliament.
At this event, Independent Age and ILC-UK, supported by members of the Ready for Ageing Alliance, launched 2030 Vision: Making the UK the best country to grow old in, which will look to the long term and consider what politicians and policy makers need to now, both in preparation for next year’s General Election, and between 2015 and 2020, to prepare for the long term opportunities and challenges ahead.
During the debate, we invited contributions on the economic and societal implications of population ageing and the major policy decisions all the main parties face to ready the UK and its public services for dramatic population ageing.
It’s clear that our political, social and cultural approach towards old age today is already hopelessly out of date, so this event will provide Parliamentarians and stakeholders from across civil society with an opportunity to mark the first anniversary of the House of Lords’ Committee report on demographic change and look ahead, so as a society we can seize the opportunities presented by an ageing population.
This file is of the digital artwork produced by Ken. The range of Ken's work is excellent and he has left us with a body of consistantle high quality, with some really exceptional pictures.
2. Warm Up Solve. 1. 2 x + 9 x – 3 x + 8 = 16 2. –4 = 6 x + 22 – 4 x 3. + = 5 4. – = 3 x = 1 x = –13 x = 34 x = 50 2 7 x 7 7 1 9 x 16 2 x 4 1 8
3. Extension of AF4.1 Solve two-step linear equations and inequalities in one variable over the rational numbers, interpret the solution or solutions in the context from which they arose, and verify the reasonableness of the results. Also covered: AF1.1 California Standards
4. Solve. 4 x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4 x + 6 = x – 4 x – 4 x 6 = –3 x To collect the variable terms on one side, subtract 4x from both sides. Since x is multiplied by -3, divide both sides by – 3. – 2 = x 6 – 3 – 3 x – 3 =
5. You can always check your solution by substituting the value back into the original equation. Helpful Hint
6. Solve. 9 b – 6 = 5 b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9 b – 6 = 5 b + 18 – 5 b – 5 b 4 b – 6 = 18 To collect the variable terms on one side, subtract 5b from both sides. Since b is multiplied by 4, divide both sides by 4. b = 6 + 6 + 6 4 b = 24 Since 6 is subtracted from 4b, add 6 to both sides. 4 b 4 24 4 =
7. Solve. 9 w + 3 = 9 w + 7 Additional Example 1C: Solving Equations with Variables on Both Sides 3 ≠ 7 There is no solution. There is no number that can be substituted for the variable w to make the equation true. 9 w + 3 = 9 w + 7 – 9 w – 9 w To collect the variable terms on one side, subtract 9w from both sides.
8. if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint
9. Solve. 5 x + 8 = x Check It Out! Example 1A 5 x + 8 = x – 5 x – 5 x 8 = –4 x Since x is multiplied by – 4, divide both sides by – 4. – 2 = x To collect the variable terms on one side, subtract 5x from both sides. 8 – 4 – 4 x – 4 =
10. Solve. 3 b – 2 = 2 b + 12 3 b – 2 = 2 b + 12 – 2 b – 2 b b – 2 = 12 + 2 + 2 b = 14 Since 2 is subtracted from b, add 2 to both sides. Check It Out! Example 1B To collect the variable terms on one side, subtract 2b from both sides.
11. Solve. 3 w + 1 = 3 w + 8 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true. Check It Out! Example 1C 3 w + 1 = 3 w + 8 – 3 w – 3 w To collect the variable terms on one side, subtract 3w from both sides.
12. To solve more complicated equations, you may need to first simplify by combining like terms or clearing fractions. Then add or subtract to collect variable terms on one side of the equation. Finally, use properties of equality to isolate the variable.
13. Solve. 10 z – 15 – 4 z = 8 – 2 z – 15 Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides 10 z – 15 – 4 z = 8 – 2 z – 15 + 15 +15 6 z – 15 = –2 z – 7 Combine like terms. + 2 z + 2 z Add 2z to both sides. 8 z – 15 = – 7 8 z = 8 z = 1 Add 15 to both sides. Divide both sides by 8. 8 z 8 8 8 =
14. Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides Multiply by the LCD, 20. 4 y + 12 y – 15 = 20 y – 14 16 y – 15 = 20 y – 14 Combine like terms. + – = y – y 5 3 4 3 y 5 7 10 y 5 3 4 3 y 5 7 10 + – = y – 20 ( ) = 20 ( ) y 5 3 4 3 y 5 7 10 + – y – 20 ( ) + 20 ( ) – 20 ( ) = 20( y ) – 20 ( ) y 5 3 y 5 3 4 7 10
15. Additional Example 2B Continued Add 14 to both sides. – 15 = 4 y – 14 – 1 = 4 y + 14 + 14 Divide both sides by 4. 16 y – 15 = 20 y – 14 – 16 y – 16 y Subtract 16y from both sides. – 1 4 4 y 4 = – 1 4 = y
16. Solve. 12 z – 12 – 4 z = 6 – 2 z + 32 Check It Out! Example 2A 12 z – 12 – 4 z = 6 – 2 z + 32 + 12 +12 8 z – 12 = –2 z + 38 Combine like terms. + 2 z + 2 z Add 2z to both sides. 10 z – 12 = 38 10 z = 50 z = 5 Add 12 to both sides. Divide both sides by 10. 10 z 50 10 10 =
17. Multiply by the LCD, 24. 6 y + 20 y + 18 = 24 y – 18 26 y + 18 = 24 y – 18 Combine like terms. + + = y – Check It Out! Example 2B y 4 3 4 5 y 6 6 8 y 4 3 4 5 y 6 6 8 + + = y – 24 ( ) = 24 ( ) y 4 3 4 5 y 6 6 8 + + y – 24 ( ) + 24 ( ) + 24 ( ) = 24( y ) – 24 ( ) y 4 5 y 6 3 4 6 8
18. Subtract 18 from both sides. 2 y + 18 = – 18 2 y = –36 – 18 – 18 Divide both sides by 2. y = –18 26 y + 18 = 24 y – 18 – 24 y – 24 y Subtract 24y from both sides. Check It Out! Example 2B Continued – 36 2 2 y 2 =
19. Additional Example 3: Business Application Daisy’s Flowers sells a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florists' bouquets cost the same price. What is the price? Daisy’s: c = 39.95 + 2.95 r Write an equation for each service. Let c represent the total cost and r represent the number of roses. total cost is flat fee plus cost for each rose Other: c = 26.00 + 4.50 r
20. Additional Example 3 Continued 39.95 + 2.95 r = 26.00 + 4.50 r Now write an equation showing that the costs are equal. – 2.95 r – 2.95 r 39.95 = 26.00 + 1.55 r Subtract 2.95r from both sides. – 26.00 – 26.00 Subtract 26.00 from both sides. 13.95 = 1.55 r Divide both sides by 1.55. 9 = r The two bouquets from either florist would cost the same when purchasing 9 roses. 13.95 1.55 1.55 r 1.55 =
21. Additional Example 3 Continued To find the cost, substitute 9 for r into either equation. Daisy’s: The cost for a bouquet with 9 roses at either florist is $66.50. c = 39.95 + 2.95 r c = 39.95 + 2.95( 9 ) c = 39.95 + 26.55 c = 66.5 Other florist: c = 26.00 + 4.50 r c = 26.00 + 4.50( 9 ) c = 26.00 + 40.50 c = 66.5
22. Check It Out! Example 3 Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both baskets cost the same price. Marla’s: c = 22.00 + 2.25 b total cost is flat fee plus cost for each balloon Other: c = 16.00 + 3.00 b Write an equation for each service. Let c represent the total cost and b represent the number of balloons.
23. Check It Out! Example 3 Continued 22.00 + 2.25 b = 16.00 + 3.00 b Now write an equation showing that the costs are equal. – 2.25 b – 2.25 b 22.00 = 16.00 + 0.75 b Subtract 2.25b from both sides. – 16.00 – 16.00 Subtract 16.00 from both sides. 6.00 = 0.75 b Divide both sides by 0.75. 8 = b The two services would cost the same when purchasing a muffin basket with 8 balloons. 6.00 0.75 0.75 b 0.75 =
24. Lesson Quiz Solve. 1. 4 x + 16 = 2 x 2. 8 x – 3 = 15 + 5 x 3. 2(3 x + 11) = 6 x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = 6 x = –8 no solution x = 36 An orange has 45 calories. An apple has 75 calories. 1 4 1 2