Download presentation

Published byGabriella Butler Modified over 4 years ago

1
**Solving Equations with Variables on Both Sides**

Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1 Holt McDougal Algebra 1

2
**Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. –6x 4. 15 – (x – 2)**

Solve. 5. 3x + 2 = 8 6. –6x –7x + 21 2x + 3 17 – x 2 28

3
Objective Solve equations in one variable that contain variable terms on both sides.

4
Vocabulary identity

5
To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation. Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.

6
**Example 1: Solving Equations with Variables on Both Sides**

Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 To collect the variable terms on one side, subtract 5n from both sides. –5n –5n 2n – 2 = 2n = Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. n = 4

7
Check It Out! Example 1a Solve 4b + 2 = 3b. 4b + 2 = 3b To collect the variable terms on one side, subtract 3b from both sides. –3b –3b b + 2 = 0 – 2 – 2 b = –2

8
Check It Out! Example 1b Solve y = 0.7y – 0.3. To collect the variable terms on one side, subtract 0.3y from both sides. y = 0.7y – 0.3 –0.3y –0.3y = 0.4y – 0.3 Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. = 0.4y Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication. 2 = y

9
To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

10
**Example 2: Simplifying Each Side Before Solving Equations**

Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 4 – 6a + 4a = –1 –5(7 – 2a) Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – a 4 – 2a = – a Combine like terms. Since –36 is added to 10a, add 36 to both sides. 40 – 2a = a + 2a a To collect the variable terms on one side, add 2a to both sides. = a

11
Example 2 Continued Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 40 = 12a Since a is multiplied by 12, divide both sides by 12.

12
Check It Out! Example 2a Solve Distribute to the expression in parentheses. 1 2 To collect the variable terms on one side, subtract b from both sides. 1 2 3 = b – 1 Since 1 is subtracted from b, add 1 to both sides. 4 = b

13
Check It Out! Example 2b Solve 3x + 15 – 9 = 2(x + 2). 3x + 15 – 9 = 2(x + 2) Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 Combine like terms. –2x –2x To collect the variable terms on one side, subtract 2x from both sides. x + 6 = – – 6 Since 6 is added to x, subtract 6 from both sides to undo the addition. x = –2

14
**An identity is an equation that is true for all values of the variable**

An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. Some equations are always false. These equations have no solutions.

15
**Identities and False Equations**

WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2 + 1 = 2 + 1 3 = 3 ALGEBRA 2 + x = 2 + x –x –x 2 = 2

16
**Identities and False Equations**

When solving an equation, if you get a false equation, the original equation has no solutions. WORDS x = x + 3 –x –x 0 = 3 1 = 1 + 2 1 = 3 ALGEBRA NUMBERS Identities and False Equations

17
**Example 3A: Infinitely Many Solutions or No Solutions**

Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 10 – 5x + 1 = 7x + 11 – 12x Identify like terms. 11 – 5x = 11 – 5x Combine like terms on the left and the right. + 5x x Add 5x to both sides. = 11 True statement. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

18
**Example 3B: Infinitely Many Solutions or No Solutions**

Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 12x – 3 + x = 5x – 4 + 8x Identify like terms. 13x – 3 = 13x – 4 Combine like terms on the left and the right. –13x –13x Subtract 13x from both sides. –3 = –4 False statement. The equation 12x – 3 + x = 5x – 4 + 8x is a false equation. There is no value of x that will make the equation true. There are no solutions.

19
** Check It Out! Example 3a Solve 4y + 7 – y = 10 + 3y.**

Identify like terms. 3y + 7 = 3y + 10 Combine like terms on the left and the right. –3y –3y Subtract 3y from both sides. 7 = False statement. The equation 4y + 7 – y = y is a false equation. There is no value of y that will make the equation true. There are no solutions.

20
**Check It Out! Example 3b Solve 2c + 7 + c = –14 + 3c + 21.**

Identify like terms. 3c + 7 = 3c + 7 Combine like terms on the left and the right. –3c –3c Subtract 3c both sides. 7 = True statement. The equation 2c c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

21
Example 4: Application Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be? Person Bulbs Jon 60 bulbs plus 44 bulbs per hour Sara 96 bulbs plus 32 bulbs per hour

22
**Example 4: Application Continued**

Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs 32 bulbs each hour When is ? b = b b = b To collect the variable terms on one side, subtract 32b from both sides. – 32b – 32b b = 96

23
**Example 4: Application Continued**

b = 96 Since 60 is added to 12b, subtract 60 from both sides. – – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

24
**Example 4: Application Continued**

After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3: b = (3) = = 192 b = (3) = = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.

25
**Let g represent Greg's age, and write expressions for his age.**

Check It Out! Example 4 Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Let g represent Greg's age, and write expressions for his age. three times Greg's age four times Greg's age is equal to increased by decreased by 7 . 3 4g – = g

26
**Check It Out! Example 4 Continued**

4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. –3g –3g g – 3 = Since 3 is subtracted from g, add 3 to both sides. g = Greg is 10 years old.

27
**Lesson Quiz 1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4**

Solve each equation. 1. 7x + 2 = 5x (2x – 5) = 5x + 4 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. 6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same? 3 8 all real numbers 1 20 hours

Similar presentations

Presentation is loading. Please wait....

OK

Properties of Real Numbers

Properties of Real Numbers

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on bermuda triangle for class 9 Slideshare net download ppt on pollution Electronics ieee papers ppt on sustainable buildings Ppt on job satisfaction Ppt on rockets and satellites Gi system anatomy and physiology ppt on cells Ppt on crash fire tender picture Ppt on periodic table of elements File type ppt on cybercrime laws Ppt on non destructive testing