2-4 Warm Up Lesson Presentation Lesson Quiz
Bell Quiz 2-4 1 pt Simplify. 1. 4x 10x 4 pts Solve the equation. 2. 3x + 2 = 8 5 pts possible
Questions on 2-3
Objective Solve equations in one variable that contain variable terms on both sides.
identity contradiction Vocabulary
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.
Example 1: Variables on Both Sides Solve 7n 2 = 5n + 6. (To save time, we won t check work on lecture notes. Still need to check for assignments.) To collect the variable terms on one side, subtract 5n from both sides. Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
Check It Out! Example 1 Solve 1a: 4b + 2 = 3b 1b: 0.5 + 0.3y = 0.7y 0.3
To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.
3 STEPS TO SOLVING (Write in notes) 1. SimPlify = PEMDAS (LHS/RHS) 2. Collect variables 3. Solve = SADMEP
Solve 4 6a + 4a = 1 5(7 2a). Distribute 5 to the expression in parentheses. Example 2A: Simplifying Each Side Before Solving Equations Combine like terms. To collect the variable terms on one side, add 2a to both sides. Add 36 to both sides. Divide both sides by 12.
Check It Out! Example 2 Solve 2a: 2b: 3x + 15 9 = 2(x + 2)
An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. A contradiction is an equation that is not true for any value of the variable. It has no solutions.
Identities and Contradictions WORDS NUMBERS ALGEBRA 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. 2 + 1 = 2 + 1 3 = 3 2 + x = 2 + x x x 2 = 2
Identities and Contradictions WORDS NUMBERS ALGEBRA Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. 1 = 1 + 2 1 = 3 x = x + 3 x x 0 = 3
Example 3A: Infinitely Many Solutions or No Solutions Solve 10 5x + 1 = 7x + 11 12x. Identify like terms. Combine like terms on the left and the right. Add 5x to both sides. 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.
Example 3B: Infinitely Many Solutions or No Solutions Solve 12x 3 + x = 5x 4 + 8x. Identify like terms. Combine like terms on the left and the right. Subtract 13x from both sides. False statement. The equation 12x 3 + x = 5x 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.
Check It Out! Example 3 Solve 3a: 4y + 7 y = 10 + 3y 3b: 2c + 7 + c = 14 + 3c + 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 Jon Sara Bulbs 60 bulbs plus 44 bulbs per hour 96 bulbs plus 32 bulbs per hour
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 plus 32 bulbs each hour When is?
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.
HOMEWORK Sec 2-4: (Pg 103) 1, 2, 8, 10, 12, 14, 15-32 all, 60, 74-77, 82, 83