More on Solving Equations

Slides:



Advertisements
Similar presentations
Chapter 1 Lesson 1 Variables and Expressions
Advertisements

Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.5 An Introduction to Problem Solving Copyright © 2013, 2009, 2006 Pearson Education,
Coordinate Algebra Practice EOCT Answers Unit 1. #1 Unit 1 A rectangle has a length of 12 m and a width of 400 cm. What is the perimeter of the rectangle?
Equations, Inequalities and Problem Solving
Additional Example 1: Solving Equations That Contain Like Terms
Unit 12 INTRODUCTION TO ALGEBRA. 2 ALGEBRAIC EXPRESSIONS An algebraic expression is a word statement put into mathematical form by using variables, arithmetic.
TRANSLATING Word Phrases to Algebraic Expressions Writing Equations
Student Learning Goal Chart Chapter 10 Pre-Algebra Learning Goal Students will understand solving linear equations and inequalities.
ALGEBRAIC WORD PROBLEMS
Today, I will learn the formula for finding the area of a rectangle.
Applications of Geometry Example 1: The perimeter of a rectangular play area is 336 feet. The length is 12 feet more than the width. Determine the dimensions.
Equations Ex. 1) 3(2x + 5) = Distribution 6x + 15 = -9 – 15 6x = x = Subtract 15 from both sides. 3. Divide both sides by 6.
12-1 Solving Two-Step Equations Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation.
More on Solving Equations Section 4. Solve: 7x + x – 2x + 9 = 15 Answer: 7x + x – 2x + 9 = 15 6x + 9 = ___________ 6x = 6 x = 1.
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
9.1 – Solving Equations.
Section 3.6 Writing Equations for Story Problems.
2.1 – Linear Equations in One Variable
 Sometimes you have a formula and you need to solve for some variable other than the "standard" one. Example: Perimeter of a square P=4s It may be that.
1.4 Solving Equations ●A variable is a letter which represents an unknown number. Any letter can be used as a variable. ●An algebraic expression contains.
Section 3.4: Solving Equations with Variables on Both Sides Algebra 1 Ms. Mayer.
PRESENTATION 11 What Is Algebra. ALGEBRAIC EXPRESSIONS An algebraic expression is a word statement put into mathematical form by using variables, arithmetic.
Solving Linear Equations Unit Test Review Game
Solving Equations & Inequalities Word Problems
VARIABLES & EXPRESSIONS Section 1.1. Quantity Anything that can be measured or counted.
$100 $200 $300 $400 $500 $200 $300 $400 $500 Problem Solving Multi-Step Equations Simple and Compound Interest Multi-Step Inequalities Transforming.
Solving Equations with Variables on Both Sides Module 4 Lesson 4.
Contents Lesson 7-1Solving Equations with Variables on Each Side Lesson 7-2Solving Equations with Grouping Symbols Lesson 7-3Inequalities Lesson 7-4Solving.
Notes Over 1.5 Write the phrase as a variable expression. Let x represent the number. 1. The sum of 1 and a number sum add switch 2. 4 less than a number.
EXAMPLE 1 Evaluate powers a. (–5) 4 b. –5 4 = (–5) (–5) (–5) (–5)= 625 = –( )= –625.
Jeopardy Solving Equations Add and Subtract Multiply and Divide Multi-Step Variables on each side Grouping Symbols $100 $200 $300 $400 $500 $100 $200.
Writing & Solving Equations
Solve Multi-Step Equations
Warm-up Solve the first system of equations by the Substitution Method, then graphing.
One step equations Add Subtract Multiply Divide  When we solve an equation, our goal is to isolate our variable by using our inverse operations.  What.
The length of a rectangle is 6 in. more than its width. The perimeter of the rectangle is 24 in. What is the length of the rectangle? What we know: Length.
4.8 Polynomial Word Problems. a) Define the variable, b) Write the equation, and c) Solve the problem. 1) The sum of a number and its square is 42. Find.
Translating Words into Symbols Objective – To translate phrases into variable expressions Words that mean addition: Sum, increased by, more than, added.
 Solve the following…  6 = x + 2  4 = q + 13  23 = b - 19  5b = 145  -7y = 28  2/3q = 18  1/5x = 2/7.
Math 71 Chapter 1. Use the order of operations to simplify each expression. What is your first step? and/or
Chapter 1 Section 3. Example 3-1a Write an algebraic expression to represent 3 more than a number. Answer:
Solve the following word problem. Manny is two years older Enrique. The sum of the their ages is 40. How old is Manny and Enrique? Let: m = Manny’s age.
Solving and Graphing Inequalities CHAPTER 6 REVIEW.
OBJECTIVE: TSW SOLVE MULTI-STEP EQUATIONS. SEPTEMBER 21,2011 Lesson 2-3 Multi- Step Equations.
11-1 Solving Two-Step Equations Do Now 1. 9b + 8 = –10 = c Solve the equation. Check you answer y – 3 = x = -16 b = 8 x.
Solving 2 step equations. Two step equations have addition or subtraction and multiply or divide 3x + 1 = 10 3x + 1 = 10 4y + 2 = 10 4y + 2 = 10 2b +
Lesson 5.1/5.2 – Writing Expressions and Equations Write this TITLE down on your notes!!! 5.1 /5.2 Writing Expressions and Equations.
Solving Equations with Variables on Both Sides. Review O Suppose you want to solve -4m m = -3 What would you do as your first step? Explain.
Lesson Days Equations and Problem Solving Pages
Jeopardy Solving Equations
Solving Two-Step Equations
Homework # 9 – Word Problems
simplify radical expressions involving addition and subtraction.
Rewrite a formula with three variables
Solving Two-Step Equations
Linear Equations Mr. HUYNH
You will have a limited amount of time to solve the following problems with a partner. Remember to DEFINE your variable first!! Good luck!
Solving Two Step Equations 11-1
2-4 Solving Multi-Step Equations
Solving Linear Equations Unit Test Review Game
Constructing Equations
Equations and Problem Solving
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Do Now b = 8 x = -4 c = -5 y = 3 Solve the equation. Check you answer.
Adding and Subtracting Radicals
By- Sabrina,Julianna, and Killian
Translating Words to Symbols
Warm Up Problem of the Day Lesson Presentation Lesson Quizzes.
Area = l(w) 38 = 2(3x + 4) 38 = 6x = 6x = x CHECK: 38 = 2(3(5) + 4) 38 = 2(15 + 4) 38 = 2(19) 38 = 38 √
Presentation transcript:

More on Solving Equations

Grouping Symbols on Both Sides…… You might have to distribute before you combine like terms.

Solve: 2(x+5)= 3(x + 2) + x Answer: 2x + 10 = 3(x + 2) + x __________________ -2x + 10 = 6 -10 -10 -2x = -4 x = 2 2(x+5)= 3(x + 2) + x

Solve: 2(3x + 2) = 2(x+8) Answer: 2(3x + 2) = 2x + 16 6x + 4 = 2x + 16 ________________ 4x + 4 = 16 -4 -4 _________________ 4x = 12 x = 3 2(3x + 2) = 2(x+8)

Word Problems

Words that tell you to add…. Plus Increased By Sum More Than

Words that tell you to subtract… Minus Decreased By Difference Less Than

Multiply and Divide Words…..

Write the mathematical statement for the following….. 9 increased by 2 5 decreased by m Sum of 4 and 2x 7 less than 12 7 increased by 5 times a number 8 decreased by 3 times a number The sum of twice a number and 5 9 + 2 5 – m 4 + 2x 12 – 7 7 + 5n 8 – 3x 2y + 5

You Try….. The sum of 7 and 3 times a number 4 times Harry’s age increased by 2 8 pounds less than twice wanda’s weight 7 + 3n 4h + 2 2w - 8

Word Problems…… The $500 selling price of a TV is $70 less than 3 times the cost. Find the profit Let’s define the variables first: Let C = Cost Let P = Profit Let SP = Selling Price

The $500 selling price of a TV is $70 less than 3 times the cost The $500 selling price of a TV is $70 less than 3 times the cost. Find the profit. Think: Cost + Profit = Selling Price C + P = SP We know the selling price and are looking for the profit. Somehow we need to find the cost first: 3C – 70 = $500 3C = $570 Cost = $190. We are still looking for the profit.

Cost + Profit = Selling Price. 190 + P = 500 P = 500 – 190 The $500 selling price of a TV is $70 less than 3 times the cost. Find the profit. We know that the cost = 190. Cost + Profit = Selling Price. 190 + P = 500 P = 500 – 190 Profit = $310 (This is the answer)

You Try…… The $140 selling price of a game is $60 less than twice the cost. Find the profit.

The $140 selling price of a game is $60 less than twice the cost The $140 selling price of a game is $60 less than twice the cost. Find the profit. 1st find the cost: 2C – 60 = 140 2C = 140 + 60 2C = 200 C = $100 2nd find the profit: C + P = SP 100 + P = 140 P = 140 – 100 Profit = $40

Example…… Mr. Daniel’s family rented a car when they flew to Hawaii for a 3-day vacation. They paid $42 per day and $0.07 for each mile driven. How much did it cost to rent the car for 3 days and drive 200 miles?

Answer….. Mr. Daniel’s family rented a car when they flew to Hawaii for a 3-day vacation. They paid $42 per day and $0.07 for each mile driven. How much did it cost to rent the car for 3 days and drive 200 miles? Total Cost = (Cost Per Day)(# of Days) + (Cost Per Mile)(# of Miles) Cost = (42)(3) + (200)(0.07) Cost = $126 + $14 Cost = $140

Answer…… A car repair shop charged Mr. Jacobs $96 for an automotive part plus $72 per hour that a mechanic worked to install the part. The total charge was $388. For about how long did the mechanic work to install the part on Mr. Jacob’s car? Total Cost = Base fee + (Charge Per Hour)(# of hours) $388 = $96 + $72(hours) 388 = 96 + 72H 388 – 96 = 72H 292 = 72H H = 4.06 or 4 Hours

Example…… A car repair shop charged Mr. Jacobs $96 for an automotive part plus $72 per hour that a mechanic worked to install the part. The total charge was $388. For about how long did the mechanic work to install the part on Mr. Jacob’s car?

Example…… The length of a rectangle is 5 and the area is 35. Find the perimeter of the rectangle.

Answer…… The length of a rectangle is 5 and the area is 35. Find the perimeter of the rectangle. 1st: Find the width by using the area formula of a rectangle. A = length x width 35 = 5 x width width = 7

Remember the width = 7…. The length of a rectangle is 5 and the area is 35. Find the perimeter of the rectangle. 2nd: Plug into the formula to find the perimeter. P = 2l + 2w P = 2(5) + 2(7) P = 10 + 14 P = 24

Example…… The perimeter of a rectangle is 20 and the width is 4. Find the area of the rectangle.

Answer….. The perimeter of a rectangle is 20 and the width is 4. Find the area of the rectangle. 1st: Find the length using the perimeter formula. P = 2w + 2l 20 = 2(4) + 2l 20 = 8 + 2l 12 = 2l l = 6

Remember the length = 6….. The perimeter of a rectangle is 20 and the width is 4. Find the area of the rectangle. 2nd: Find the area. A = l x w A = 6 x 4 A = 24