Solving Equations Objectives: 1)To solve equations 2)To solve problems by writing equations.

Slides:



Advertisements
Similar presentations
Chapter 1 Tools of Algebra.
Advertisements

Summary Subsets of Real Numbers
Chapter 1: Tools of Algebra 1-3: Solving Equations Essential Question: What is the procedure to solve an equation for a variable?
( ) EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1
Solving Equations by Factoring
EXAMPLE 2 Use a ratio to find a dimension SOLUTION Painting You are planning to paint a mural on a rectangular wall. You know that the perimeter of the.
2.5Formulas and Additional Applications from Geometry
7.2, 7.3 Solving by Substitution and Elimination
EXAMPLE 1 Finding Perimeter and Area SOLUTION Find the perimeter. P = 2l + 2w Write formula. = 2 ( 8 ) + 2( 5 ) Substitute. = 26 Multiply, then add. Find.
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.
EXAMPLE 1 Collecting Like Terms x + 2 = 3x x + 2 –x = 3x – x 2 = 2x 1 = x Original equation Subtract x from each side. Divide both sides by x2x.
System of linear inequalities A system of linear inequalities is made up of two or more inequalities A solution of a System of linear inequalities is.
Solve equations that involve grouping symbols
A solution of a system of two equations in two variables is an ordered pair of numbers that makes both equations true. A solution to two equations (1,
6.1 – Ratios, Proportions, and the Geometric Mean Geometry Ms. Rinaldi.
Bell Ringer.
Algebra 2 Lesson 1-3 (Page 18) ALGEBRA 2 LESSON 1-3 Solving Equations 1-1.
Chapter 3 Lesson 7 Using Formulas pgs What you will learn: Solve problems by using formulas Solve problems involving the perimeters & areas of.
Solving systems of equations with 2 variables
Chapter 10 Section 3 Solving Quadratic Equations by the Quadratic Formula.
Ratio and Proportion 7-1.
6.1 – Ratios, Proportions, and the Geometric Mean.
1.3 Solving Equations Properties Reflexive: Symmetric: Transitive: Substitution: If a = b, then b can be substituted anywhere for a.
 Objectives: 1. To solve equations 2. To solve problems by writing equation.
Unit 1: Perimeter and Area. MFM1P Lesson 1: Perimeter and Area of Basic Shapes Learning Goals: I can determine the perimeter of a variety of shapes I.
Lesson 5.6-Number Problems Obj: To solve # word problems.
Chapter 2 Section 5. Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Formulas and Additional Applications from Geometry Solve a formula.
EXAMPLE 2 Use a ratio to find a dimension SOLUTION Painting You are planning to paint a mural on a rectangular wall. You know that the perimeter of the.
Solving Equations by Factoring Definition of Quadratic Equations Zero-Factor Property Strategy for Solving Quadratics.
Chapter 6 Similarity Pre-Requisite Skills Page 354 all.
Optimization Problems Example 1: A rancher has 300 yards of fencing material and wants to use it to enclose a rectangular region. Suppose the above region.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 2 Equations, Inequalities and Problem Solving.
Chapter 2 Section 1 Relations and Functions Algebra 2 Notes January 8, 2009.
Section 5Chapter 6. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 3 Solving Equations by Factoring Learn and use the zero-factor.
Chapter 6 Section 1 - Slide 1 Copyright © 2009 Pearson Education, Inc. Chapter 6 Section 1 - Slide 1 1. Algebra 2. Functions.
Quiz 1-3 Quiz Solve for x: 3. Simplify: 4. What property is illustrated below?
Algebra 2 Lesson 1-3 Examples (part 2). Solving Equations A solution to an equation is __________________________________________ __________________________________________.
Lesson 1-3 Solving Equations. Properties of Equality Reflexive Propertya = a Symmetric PropertyIf a = b, then b = a. Transitive PropertyIf a = b and b.
Copyright 2013, 2010, 2007, 2005, Pearson, Education, Inc.
Lesson 1-3 Solving Multi-Step Equations
Finding Perimeter and Area
7.1 OBJ: Use ratios and proportions.
Solving Equations by Factoring
Copyright © 2006 Pearson Education, Inc
System of linear inequalities
Find the perimeter of the figure
CHAPTER 3 SECTION 7.
Solving Rational Equations
Steps to solving a word problem
Solving Multi-Step Equations
Solving Multi-Step Equations
Obj: to use ratios to solve problems
ZPP Zero Product Property If AB = 0 then A = 0 or B = 0.
2.5 Formulas and Additional Applications from Geometry
Use the substitution method to find all solutions of the system of equations {image} Choose the answer from the following: (10, 2) (2, 10) (5, - 2) ( -
Ratio Ratio – a comparison of numbers A ratio can be written 3 ways:
In all things of nature there is something of the marvelous.
Clear fractions and decimals
Problem Solving and Using Formulas
 
Standard Form Quadratic Equation
Rewrite Equations and Formulas
Find the perimeter of the figure
one of the equations for one of its variables Example: -x + y = 1
2 Equations, Inequalities, and Applications.
Area & Perimeter.
Algebra 1 Section 12.3.
5-Minute Check Solve each equation. Check your solution. x/3 =
Chapter 3 Section 6 Applications.
Presentation transcript:

Solving Equations Objectives: 1)To solve equations 2)To solve problems by writing equations

Solution to An Equation A number that makes the equation true is a solution to the equation. Ex: 5x + 23 = x = x = 13 5(13) + 23 = = = 88 Since x = 13 satisfies as an answer to 5x + 23 = 88, we call 13 the solution Now, check your answer

Example #2: Using the Distributive Property Solve 3x – 7(2x – 13) = 3(-2x + 9) 3x – 14x + 91= -6x x + 91 = -6x x +11x 91 = 5x = 5x = x

Example #3: Solving a Formula for One of its Variables The formula for the area of a trapezoid is A = ½ h(b 1 +b 2 ). Solve the formula for h 2(A) = 2( ½) h(b 1 +b 2 ) 2A = h (b 1 +b 2 ) (b 1 +b 2 ) 2A/ (b 1 +b 2 ) = h

Example #4: Solving an Equation for One of its Variables Solve x/a + 1 = x/b for any restrictions on a & b ab(x/a) + (ab)(1) = (ab)(x/b) bx + ab = ax ab = ax – bx ab = (a – b)x (a-b) ab/(a-b) = x

Example #5 Word Problems A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. 1 st, remember the formula for the perimeter of a rectangle. P = 2w + 2l or P = 2(w + l) Since the length is 5 times the width, we’ll represent everything in terms of width. w = width 5w = length

Example #5 Word Problems Write the equation: 2w + 2(5w) = 100 2w + 10w = w = w = 8 1/3 Remember that w represents the width, so 5w = 41 2/3 is the length

Example #6: Using Ratios The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18in. Find the lengths of the sides. Use the ratio and the perimeter to create an equation: 3x + 4x + 5x = 18 12x =

Example #6: Using Ratios x = 1.5 Now that you know what x is, it’s time to find 3x, 4x, 5x 3x = 3(1.5) = 4.5 4x = 4(1.5) = 6 5x = 5(1.5) = 7.5

Example #7: Word Problems Radar detected an unidentified plane 5000mi away, approaching at 700mi/h. Fifteen minutes later an interceptor plane was dispatched, traveling at 800mi/h. How long did the interceptor take to reach the approaching plane?

Example #7: Word Problems Find a relationship in the information. Distance for approaching plane + distance for the interceptor = 5000mi Let t = the time in hours for the interceptor t = the time in hours for the approaching plane. 800 t + 700(t ) = 5000

Example #7: Word Problems 800 t (t ) = t +700t = t = t = t = or 3hr 13min