Download presentation

Presentation is loading. Please wait.

Published byCorey Jones Modified over 4 years ago

1
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. Therefore, it is often easiest to move the term with the smaller coefficient first.

2
Solve 7n – 2 = 5n + 6. Example 1: Solving Equations with Variables on Both Sides To collect the variable terms on one side, subtract 5n from both sides. 7n – 2 = 5n + 6 –5n 2n – 2 = 6 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. 2n = 8 + 2 n = 4

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

4
Solve 0.5 + 0.3y = 0.7y – 0.3. Check It Out! Example 1b To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.

5
To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

6
Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2: Simplifying Each Side Before Solving Equations Combine like terms. Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +36 40 – 2a = 10a + 2a +2a 40 = 12a Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides.

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

8
Solve. Check It Out! Example 2a Since 1 is subtracted from b, add 1 to both sides. Distribute to the expression in parentheses. 1 2 + 1 3 = b – 1 To collect the variable terms on one side, subtract b from both sides. 1 2 4 = b

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

10
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.

11
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 Identities and False Equations

12
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

13
Solve 10 – 5x + 1 = 7x + 11 – 12x. Example 3A: Infinitely Many Solutions or No Solutions Add 5x to both sides. Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x 11 = 11 + 5x + 5x True statement. Combine like terms on the left and the right. 10 – 5x + 1 = 7x + 11 – 12x 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.

14
Solve 12x – 3 + x = 5x – 4 + 8x. Example 3B: Infinitely Many Solutions or No Solutions Subtract 13x from both sides. Identify like terms. 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –3 = –4 –13x False statement. Combine like terms on the left and the right. 12x – 3 + x = 5x – 4 + 8x 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.

15
Solve 4y + 7 – y = 10 + 3y. Check It Out! Example 3a Subtract 3y from both sides. Identify like terms. 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 7 = 10 –3y –3y False statement. Combine like terms on the left and the right. 4y + 7 – y = 10 + 3y The equation 4y + 7 – y = 10 + 3y is a false equation. There is no value of y that will make the equation true. There are no solutions.

16
Solve 2c + 7 + c = –14 + 3c + 21. Check It Out! Example 3b Subtract 3c both sides. Identify like terms. 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 7 = 7 –3c –3c True statement. Combine like terms on the left and the right. 2c + 7 + c = –14 + 3c + 21 The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

17
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? Example 4: Application PersonBulbs Jon60 bulbs plus 44 bulbs per hour Sara96 bulbs plus 32 bulbs per hour

18
Example 4: Application Continued Let b = bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour When is ? 60 + 44b = 96 + 32b – 32b To collect the variable terms on one side, subtract 32b from both sides. 60 + 12b = 96

19
Example 4: Application Continued Since 60 is added to 12b, subtract 60 from both sides. 60 + 12b = 96 –60 – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

20
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: 60 + 44b 60 + 44(3) 60 + 132 192 96 + 32b 96 + 32(3) 96 + 96 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.

21
Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Check It Out! Example 4 Let g = Greg's age, and write expressions for his age. four times Greg's age decreased by 3 is equal to three times Greg's age increased by 7. 4g – 3 = 3g + 7

22
Check It Out! Example 4 Continued 4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. g – 3 = 7 –3g Since 3 is subtracted from g, add 3 to both sides. + 3 + 3 g = 10 Greg is 10 years old.

23
Homework Assignment Sec. 1-5 Wksht

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google