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Prerequisite Skills Review. Before Chapter 3. Finding Percents of Numbers. Finding Percents of Numbers 36% of 94.5. Finding Percents of Numbers. Finding Percents of Numbers 36% of 94.5 (0.36)(94.5) = 34.02. Finding Percents of Numbers. Find the percent of the number. 16% of 540.
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Prerequisite Skills Review Before Chapter 3
Finding Percents of Numbers Finding Percents of Numbers 36% of 94.5
Finding Percents of Numbers Finding Percents of Numbers 36% of 94.5 (0.36)(94.5) = 34.02
Finding Percents of Numbers Find the percent of the number. 16% of 540
Finding Percents of Numbers Find the percent of the number. 29% of 67.3
Finding Percents of Numbers Find the percent of the number. 4% of 15
Checking Possible Solutions Check whether the number 2 is a solution of the equation:
Checking Possible Solutions Check whether the number 2 is a solution of the equation: NO!!
Checking Possible Solutions Check whether -3 is a solution of the equation:
Checking Possible Solutions Check whether 9 is a solution of the equation:
Using the Distributive Property Use the Distributive Property to simplify the expression.
Using the Distributive Property Use the Distributive Property to simplify the expression.
Using the Distributive Property Use the Distributive Property to simplify the expression.
Using the Distributive Property Use the Distributive Property to simplify the expression.
Using the Distributive Property Use the Distributive Property to simplify the expression.
Simplifying Like Terms Find the terms of the expression. Then simplify by combining like terms.
Simplifying Like Terms Find the terms of the expression. Then simplify by combining like terms.
Simplifying Like Terms Find the terms of the expression. Then simplify by combining like terms.
Simplifying Like Terms Find the terms of the expression. Then simplify by combining like terms.
Chapter 3 Solving Linear Equations 3.1. Solving Equations Using Addition and Subtraction
You find $10 on the sidewalk and put it in your pocket. Later that day you empty your pocket and find that you have $28. How much did you have before you find the $10?
You could find the answer by using the model: x + 10 = 28 or 28 - 10 = x
1. Open your Book to Page 133.2. Look at Example 1. x – 5 = -133. Why is 5 instead of 13 added to both sides of the equation to solve for x?
What is a “solution step”? Each time you apply a transformation to an equation, you are writing a solution step. Solution steps are written one below the other with the equals signs aligned.
What is a “Linear Equation”? In a linear equation the variable is raised to the first power and does not occur in a denominator, inside a square root symbol, or inside absolute value symbols. Linear Equation Not a Linear Equation
What is a “Linear Equation”? In Chapter 4, you will see that Linear Equations get their name from the fact that their graphs are straight lines.
Translating Verbal Statements Match the real-life problem with an equation: You owe $16 to your cousin. You paid x dollars back and you now owe $4. How much did you pay back? x – 4 = 16 x = 16 + 4 16 – x = 4
Translating Verbal Statements Match the real-life problem with an equation: x + 16 = 4 x = 16 + 4 16 – x = 4 The temperature was xo F. It rose 16o F and is now 4oF. What was the original temperature?
Translating Verbal Statements Match the real-life problem with an equation: A telephone pole extends 4 feet below ground and 16 feet above ground. What is the total length x of the pole? x – 4 = 16 x = 16 + 4 16 – x = 4
PRACTICE Solve: x – 9 = -17
PRACTICE Solve: x – 9 = -17 x = -8
PRACTICE Solve: – 11 = n – (-2)
PRACTICE Solve: – 11 = n – (-2) n = -13
Practice The normal high temperature in January in Bismarck, North Dakota, is 20oF and the normal low temperature is -2oF. How many degrees apart are the normal high and low temperatures?
Practice The normal high temperature in January in Bismarck, North Dakota, is 20oF and the normal low temperature is -2oF. How many degrees apart are the normal high and low temperatures? 22oF
3.1. Closure Question Describe in words how you would solve the equation x + 6 = -4 for x using inverse operations.
3.1. Closure Question Describe in words how you would solve the equation x + 6 = -4 for x using inverse operations. Subtract 6 from both sides of the equation to get x = -10.
Chapter 3 Solving Linear Equations 3.2. Solving Equations Using Multiplication and Division
Properties of Equality The transformations used to isolate the variable in Lessons 3.1 and 3.2 are based on rules of algebra called “properties of equality.”