Math 3201 Notes Chapter 4: Rational Expressions & Equations

Similar documents
Ch 9.3 Geometric Sequences and Series Lessons

Here are the coefficients of the terms listed above: 3,5,2,1,1 respectively.

South Slave Divisional Education Council. Math 10C

Math 10C Long Range Plans

Polynomial Functions and Models. Learning Objectives. Polynomials. P (x) = a n x n + a n 1 x n a 1 x + a 0, a n 0

Arithmetic Sequences

1.2 Binomial Coefficients and Subsets

9.1. Sequences and Series. Sequences. What you should learn. Why you should learn it. Definition of Sequence

FURTHER INTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)

OCR Statistics 1. Working with data. Section 3: Measures of spread

Math Section 2.2 Polynomial Functions

12-5A. Equivalent Fractions and Decimals. 1 Daily Common Core Review. Common Core. Lesson. Lesson Overview. Math Background

NTH, GEOMETRIC, AND TELESCOPING TEST

CSC165H1 Worksheet: Tutorial 8 Algorithm analysis (SOLUTIONS)

SEQUENCES AND SERIES

1. The lines intersect. There is one solution, the point where they intersect. The system is called a consistent system.

9 x and g(x) = 4. x. Find (x) 3.6. I. Combining Functions. A. From Equations. Example: Let f(x) = and its domain. Example: Let f(x) = and g(x) = x x 4

Alpha Individual Solutions MAΘ National Convention 2013

1.8 What Comes Next? What Comes Later?

It just came to me that I 8.2 GRAPHS AND CONVERGENCE

The isoperimetric problem on the hypercube

MATHEMATICAL METHODS OF ANALYSIS AND EXPERIMENTAL DATA PROCESSING (Or Methods of Curve Fitting)

4.1: Equivalent and Rational Expressions

Numerical Methods Lecture 6 - Curve Fitting Techniques

condition w i B i S maximum u i

CIS 121 Data Structures and Algorithms with Java Fall Big-Oh Notation Tuesday, September 5 (Make-up Friday, September 8)

The number n of subintervals times the length h of subintervals gives length of interval (b-a).

Recursive Procedures. How can you model the relationship between consecutive terms of a sequence?

Lecture 1: Introduction and Strassen s Algorithm

Ones Assignment Method for Solving Traveling Salesman Problem

Civil Engineering Computation

Chapter 3 Classification of FFT Processor Algorithms

An (or ) is a sequence in which each term after the first differs from the preceding term by a fixed constant, called the.

SAMPLE VERSUS POPULATION. Population - consists of all possible measurements that can be made on a particular item or procedure.

Counting Regions in the Plane and More 1

Module 8-7: Pascal s Triangle and the Binomial Theorem

EVALUATION OF TRIGONOMETRIC FUNCTIONS

Lecturers: Sanjam Garg and Prasad Raghavendra Feb 21, Midterm 1 Solutions

Computer Science Foundation Exam. August 12, Computer Science. Section 1A. No Calculators! KEY. Solutions and Grading Criteria.

Name Date Hr. ALGEBRA 1-2 SPRING FINAL MULTIPLE CHOICE REVIEW #1

CONTINUI TY. JEE-Mathematics. Illustration 1 : Solution : Illustration 2 : 1. CONTINUOUS FUNCTIONS :

Solution printed. Do not start the test until instructed to do so! CS 2604 Data Structures Midterm Spring, Instructions:

Chapter 5. Functions for All Subtasks. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

Homework 1 Solutions MA 522 Fall 2017

Major CSL Write your name and entry no on every sheet of the answer script. Time 2 Hrs Max Marks 70

EE260: Digital Design, Spring /16/18. n Example: m 0 (=x 1 x 2 ) is adjacent to m 1 (=x 1 x 2 ) and m 2 (=x 1 x 2 ) but NOT m 3 (=x 1 x 2 )

4.3 Modeling with Arithmetic Sequences

Chapter 9. Pointers and Dynamic Arrays. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

ENGI 4421 Probability and Statistics Faculty of Engineering and Applied Science Problem Set 1 Descriptive Statistics

Algorithm Design Techniques. Divide and conquer Problem

Big-O Analysis. Asymptotics

The Closest Line to a Data Set in the Plane. David Gurney Southeastern Louisiana University Hammond, Louisiana

On (K t e)-saturated Graphs

Python Programming: An Introduction to Computer Science

15-859E: Advanced Algorithms CMU, Spring 2015 Lecture #2: Randomized MST and MST Verification January 14, 2015

n Some thoughts on software development n The idea of a calculator n Using a grammar n Expression evaluation n Program organization n Analysis

Mathematical Stat I: solutions of homework 1

Evaluation scheme for Tracking in AMI

6.854J / J Advanced Algorithms Fall 2008

Content: Learning Area: Mathematics. Integration: Grade: 9. Lesson: Module 2: Algebra and geometry. Teaching methods and Lesson Progression:

Examples and Applications of Binary Search

Optimal Mapped Mesh on the Circle

. Written in factored form it is easy to see that the roots are 2, 2, i,

CS : Programming for Non-Majors, Summer 2007 Programming Project #3: Two Little Calculations Due by 12:00pm (noon) Wednesday June

One advantage that SONAR has over any other music-sequencing product I ve worked

How do we evaluate algorithms?

CSC 220: Computer Organization Unit 11 Basic Computer Organization and Design

ENGR Spring Exam 1

CMPT 125 Assignment 2 Solutions

2) Give an example of a polynomial function of degree 4 with leading coefficient of -6

Chapter 11. Friends, Overloaded Operators, and Arrays in Classes. Copyright 2014 Pearson Addison-Wesley. All rights reserved.

A Resource for Free-standing Mathematics Qualifications

UNIT 1 RECURRENCE RELATIONS

BACHMANN-LANDAU NOTATIONS. Lecturer: Dr. Jomar F. Rabajante IMSP, UPLB MATH 174: Numerical Analysis I 1 st Sem AY

The Inverse z-transform

What are we going to learn? CSC Data Structures Analysis of Algorithms. Overview. Algorithm, and Inputs

Bezier curves. Figure 2 shows cubic Bezier curves for various control points. In a Bezier curve, only

Big-O Analysis. Asymptotics

Intro to Scientific Computing: Solutions

Creating Exact Bezier Representations of CST Shapes. David D. Marshall. California Polytechnic State University, San Luis Obispo, CA , USA

Lecture Notes 6 Introduction to algorithm analysis CSS 501 Data Structures and Object-Oriented Programming

Section 7.2: Direction Fields and Euler s Methods

Name Date Hr. ALGEBRA 1-2 SPRING FINAL MULTIPLE CHOICE REVIEW #2

Graceful Labelings of Pendant Graphs

Improving Template Based Spike Detection

IJESMR International Journal OF Engineering Sciences & Management Research

Pattern Recognition Systems Lab 1 Least Mean Squares

Normal Distributions

WebAssign Lesson 6-1b Geometric Series (Homework)

University of Waterloo Department of Electrical and Computer Engineering ECE 250 Algorithms and Data Structures

Computers and Scientific Thinking

CIS 121 Data Structures and Algorithms with Java Spring Stacks and Queues Monday, February 12 / Tuesday, February 13

IMP: Superposer Integrated Morphometrics Package Superposition Tool

Elementary Educational Computer

Chapter 2. C++ Basics. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

The Graphs of Polynomial Functions

27 Refraction, Dispersion, Internal Reflection

EE University of Minnesota. Midterm Exam #1. Prof. Matthew O'Keefe TA: Eric Seppanen. Department of Electrical and Computer Engineering

A Generalized Set Theoretic Approach for Time and Space Complexity Analysis of Algorithms and Functions

Basic allocator mechanisms The course that gives CMU its Zip! Memory Management II: Dynamic Storage Allocation Mar 6, 2000.

Transcription:

Learig Goals: See p. tet.. Equivalet Ratioal Epressios ( classes) Read Goal p. 6 tet. Math 0 Notes Chapter : Ratioal Epressios & Equatios. Defie ad give a eample of a ratioal epressio. p. 6. Defie o-permissible values as they relate to ratioal epressios. p. 7. Determie the o-permissible values for a ratioal epressio. p. 7. Defie ad give eamples of equivalet ratioal epressios. p. 8 5. Fid a ratioal epressio that is equivalet to a give ratioal epressio usig multiplicatio or divisio. p. 8 6. Determie if two ratioal epressios are equivalet. p. 0 Def A ratioal epressio is a epressio of the form p, where p ad q are polyomials ad 0 q q. It is a fractio that has a polyomial i the umerator ad a polyomial i the deomiator ad the deomiator caot be 0. E.g.: t q 7t ;, r 0;, ;, t ;, q,;, y 0, r t q q 8y No e.g.:,, 0 w If you look back at the eamples of ratioal epressios, you should otice that ratioal epressios with oe or more variables i the deomiator had restrictios o the variable(s). The restrictios o the deomiator occur because divisio by zero is udefied, so we are ot permitted to substitute certai values for the variable(s) i the deomiator because they result i divisio by zero. These value(s) which caot be substituted for the variable(s) are called o-permissible values. These opermissible values lead to the restrictios o the ratioal epressio. Def : No-permissible values are those values of the variable that makes the deomiator of the ratioal epressios equal to zero. E.g.: For, the o-permissible values are, so the restrictios are. E.g.: For are 5 5 7 0 5 7 5,. 7 See Need to Kow, bullets -, p., the o-permissible values are 5 ad, so the restrictios 7

Do # b, 9 a, 0 b, c, d, a(i) & a(ii), 6, CYU/Practisig pp. - tet i your homework booklet. Equivalet Ratioal Epressios Fidig equivalet ratioal epressios is the same as fidig equivalet umerical fractios. You must multiply (or divide) the umerator ad deomiator by the same umber or epressio. The origial ad resultig ratioal epressios are equivalet because multiplyig both the umerator ad the deomiator by the same umber or epressio meas you are multiplyig by oe. However, with ratioal epressios ivolvig variables, you have to pay close attetio to the restrictio(s) o the variable(s). E.g.: ad 6 are equivalet ratioal epressios. 8 E.g.: 5 ad 5 5 are equivalet ratioal epressios. 0 0 5 E.g.: ad are equivalet ratioal epressios for 0. E.g.: E.g.: t, 7. E.g.: 7 7 7 ad 7 5tt 5t 5t t 5t 0t ad 7 t 7 t t 7 t t 5t t t t t ad t t t are equivalet ratioal epressios for. are equivalet ratioal epressios for are equivalet ratioal epressios for t. Note: If the ratioal epressios cotai at most oe variable, you ca determie if two ratioal epressios are equivalet by graphig both epressios. If oe epressio graphs o top of the other epressio, they are equivalet. E.g.: Determie if is equivalet to for 0. The graphs of y (left graph) ad ad are equivalet. y (right graph) graph o top of each other so

Sample Eam Questio 6a Idetify the ratioal epressio that is equivalet to. a a A. ; a,0 a B. ; a,0 a C. a ; a,0 a D. 6a ; a,0 a See Key Ideas, bullet, p. Do # a, b, c,, 5,, 5, CYU/Practisig pp. - tet i your homework booklet.

. Simplifyig Ratioal Epressios ( classes) Read Goal p. 5 tet.. Simplify ratioal epressios. pp. 6-8. Determie whe a ratioal epressio is simplified. p. 5. Determie whe to idetify the o-permissible values whe simplifyig a ratioal epressio. p. 5. Idetify errors i the simplificatio of a ratioal epressio. p. 0 Simplifyig Ratioal Epressios Simplifyig ratioal epressios is very similar to simplifyig umerical fractios. The idea is to factor both the umerator ad deomiator ad cacel like factors to get a equivalet epressio. Do t forget to pay attetio to the o-permissible values so you ca state the restrictio(s) o the variable(s). Remember that graphig ca be used to check your work if there is o more tha oe variable. You have simplified a ratioal epressio whe the umerator ad the deomiator have o commo factors (see TIP, p. 5). E.g.: E.g.: 7 5 7 5 5 9r 9 r r r ; r 0 6r 6 r r r 8 6 8 8 E.g.: 8, Note that you determie the o-permissible values ad the restrictios after you factor but before you cacel (see TIP p. 5). E.g.: a a aa a a a a a ; a E.g.: E.g.:, 5 5 5 5 5 5 5 5,

6 E.g.: Simplify 6 6 6 6 6 6 E.g.: Simplify 0 8 0 6 ; 6 5 5 0 8 5 5 5 0 6 5 5 5 a a E.g.: Simplify 5a a a 5a 7a a 7a 5 a ; a 0 5 E.g.: Simplify 8 0 8 0 5 ; 0, 5 5 Sample Eam Questio ; ; 0, 5 Simplify 6 7 ANS: ; 7 5

E.g.: Fid the error i the followig simplificato of 6 0. 0 6 0 6 0 0 0 6 8; 0 The error occurs i cacelig 0 ad 0to get. The correct simplificatio is 6 0 0 80 5 80 ; 0 5 a E.g.: Fid the errors i the followig simplificato of a. a a a a a a a There are errors, oe i the simplificatio correct simplificatio is a a a a a a a a a a ; a istead of a a ad oe i the restrictios. The ; a a See Key Ideas & Need to Kow, p. 9 Do # s, b, c, d, a, c, c, d, 5 b, c, 6-8,, CYU/Practisig pp. 9- tet i your homework booklet. 6

. Multiplyig ad Dividig Ratioal Epressios ( classes) Read Goal o p. tet Outcomes:. Compare ad cotrast multiplyig ad dividig ratioal epressios with multiplyig ad dividig umerical fractios pp. -6. Determie the restrictio(s) o the variable(s) whe multiplyig or dividig ratioal epressios. pp. -6. Multiply ad divide ratioal epressios. pp. -6 Multiplyig Ratioal Epressios Multiplyig ratioal epressios is similar to multiplyig umerical fractios. However, you must watch for the restrictio(s) o the variable(s). E.g.: Simplify 5 7 We multiply across the top ad across the bottom to get 5 5 7 8 Sometimes, we ca do some cacellig before we multiply. E.g.: Simplify 5 7 5 7 5 7 5 5 5 Multiplyig ratioal epressios is similar ecept that we have to determie the o-permissible value(s) for the epressio ad use them to write the restrictios o the variable. 9 E.g.: Simplify We have to rewrite the epressio so that everythig is factored ad cacel like factors. Do t forget to fid the o-permissible values before you cacel. You could multiply across the top ad across the bottom ad the factor but this would greatly icrease the difficulty of the factorig. 9 ; 0,, 7

E.g.: Simplify ; 0, E.g.: Simplify 8 6 5 8 6 5 5 See Key Ideas & Need to Kow, p. 7 6 8 5 5 5 6 ;,0 5 Do # s a, c, a, b, 5 a, 6, CYU/Practisig pp. 8-9 tet i your homework booklet. Dividig Ratioal Epressios Dividig ratioal epressios is similar to dividig umerical fractios. However, you must watch for the restrictio(s) o the variable(s). E.g.: Simplify 5 7 First we multiply the first fractio by the reciprocal of the secod fractio ad the we multiply across the top ad across the bottom to get 5 5 7 57 5 7 Like multiplyig umerical fractios, sometimes we ca use cacellatio after we multiply by the reciprocal. E.g.: Simplify 6 5 5 6 6 5 6 5 5 5 5 5 5 5 0 Dividig ratioal epressios is similar ecept that we have to determie the o-permissible value(s) for the epressio. This is a little bit trickier tha with multiplyig ratioal epressios. 8

E.g.: Simplify t t s s t t t s t s st, s 0, t 0 s s s t 6st Do you see why t 0? E.g.: Simplify ; E.g.: Simplify, E.g.: Simplify y y y y y y y y y y y y y y y y y y y y y y y, y,0 y a a E.g.: Simplify 5 a a5 a a 5 a a5 a a 5 a 5 a a 5 a a5 5 a 5 a a a a 5 5 a 5 a a ; a, 5 5 a y y y y 9

Note that divisio ad multiplicatio of ratioal epressios ca be combied as i the eample below. E.g.: Simplify 6 6 ;, Sample Eam Questio Simplify s s 5 9 s s 5 s s5 s s 5 s 9 s s5 9s s 5 s 5 s 5 ; 5, s 6 s 9 See Key Ideas & Need to Kow, p. 7 Do # s b, d, c, d, 5 b, 7, CYU/Practisig pp. 8-9 tet i your homework booklet. 0

. Addig ad Subtractig Ratioal Epressios ( classes) Read Goal o p. tet Outcomes:. Compare ad cotrast addig ad subtractig ratioal epressios with addig ad subtractig umerical fractios pp. -8. Determie the restrictio(s) o the variable(s) whe addig or subtractig ratioal epressios. pp. -8. Add ad subtract ratioal epressios with like deomiators. pp. -8. Add ad subtract ratioal epressios with ulike deomiators. pp. -8 Addig Ratioal Epressios Addig ratioal epressios is similar to addig umerical fractios. It ivolves factorig the deomiators ad fidig the LCM of the deomiators. E.g.: Simplify 5 Step : Fid the lowest Lowest commo deomiator is. Step : Write both umerators over a sigle 5 Step : Add the umerators. 8 Step : Reduce, if ecessary 8 Sometimes, we have to fid a commo deomiator before we add. E.g.: Simplify 5 7 Step : Fid the lowest Lowest commo deomiator is. Step : Write equivalet fractios with the 5 6 Step : Write both umerators over a sigle Step : Add the umerators. 7 5 6 Step 5: Reduce, if ecessary 7 7

Addig ratioal epressios is similar ecept that we have to determie the o-permissible value(s) for the epressio so we ca write the restrictios. E.g.: Simplify m m Step : Fid the lowest Lowest commo deomiator is. Step : Write both umerators over a sigle mm Step : Combie like terms. m Step : Reduce, if ecessary m is already reduced. Step 5: Fid the o-permissible values. 0 Step 6: Write the aswer with ay restrictios m ; 0 E.g.: Simplify 0 m 8 m m m Step : Fid the lowest Lowest commo deomiator is m. Step : Write both umerators over a sigle 0m 8m m Step : Combie like terms. 8m 7 m Step : Reduce, if ecessary 8m 7 is already reduced. m Step 5: Fid the o-permissible values. m Step 6: Write the aswer with ay restrictios 8m 7 ; m m E.g.: Simplify 6 Step : Factor both deomiators. Both deomiators caot be factored.. Step : Fid the lowest Lowest commo deomiator is Step : Write both epressios with the commo 6 6 deomiator. Step : Write both umerators over a sigle 6

Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary 6 0 0 is already reduced. Step 8: Fid the o-permissible values., Step 9: Write the aswer with ay restrictios 0 ;, For may ratioal epressios, the deomiators are NOT like, so we have to fid a commo deomiator before we ca add. We may have to factor the deomiators before we ca fid the E.g.: Simplify p p Step : Factor both deomiators. p p p p p p Step : Fid the lowest Lowest commo deomiator is p p Step : Write both epressios with the commo deomiator. p p p p p p p p p p Step : Write both umerators over a sigle p p p Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary p p p p 7 p p p 7 p p is already reduced. Step 8: Fid the o-permissible values. p Step 9: Write the aswer with ay restrictios p 7 ; p p p.

E.g.: Simplify Step : Factor both deomiators. Step : Fid the lowest Lowest commo deomiator is Step : Write both epressios with the commo deomiator. Step : Write both umerators over a sigle Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary is already reduced. Step 8: Fid the o-permissible values. 0, Step 9: Write the aswer with ay restrictios ; 0,. E.g.: Simplify 0 6 5 0 Step : Factor both the umerators ad the deomiators ad cacel ay like factors. 5 5 7 5 0 7 6 0 0 0 Step : Fid the lowest Lowest commo deomiator is 7 0 Step : Write both epressios with the commo 0 0 deomiator. 7 7 0 7 0 7 7 0 0 7 7 0 Step : Write both umerators over a sigle 0 7 70.

Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary 0 70 7 70 7 70 is already reduced. Step 8: Fid the o-permissible values. 7,,5,0 Step 9: Write the aswer with ay restrictios 7 ; 7,,5,0 7 0 E.g.: Fid the error i the simplificatio below ad write the correct simplificatio. 5 5 8 The error occurs whe the deomiators are added. The correct simplificatio is 5 0 0 5 5 5 5 5 5 E.g.: Chris purchased a motorboat ad used it to travel 6 km upstream ad the 6 km back dowstream. If the river s curret is costat at km/h, algebraically determie a equatio which models the time take to travel the km. Let v be the speed of the boat i still water. Let v be the speed of the boat goig dowstream. Let v be the speed of the boat goig upstream. Upstream Dowstream d 6km d 6km Speed v Speed v 6 t t 6 v v Sice the total time is the sum of the time goig upstream ad the dowstream, the epressio for the 6 6 total time take to travel km is v v See I Summary, p. 8 Do # s,, b, d, 5 b, d, 6 a, c, 7 a, 0, CYU/Practisig pp. 8-50 tet i your homework booklet. 5

Subtractig Ratioal Epressios Subtractig ratioal epressios is similar to subtractig umerical fractios. It ivolves factorig the deomiators ad fidig the LCM of the deomiators. However, you must watch for the restrictio(s) o the variable(s). E.g.: Simplify 5 Step : Fid the lowest Lowest commo deomiator is. Step : Write both umerators over a sigle 5 Step : Subtract the umerators. Step : Reduce, if ecessary Sometimes, we have to fid a commo deomiator before we subtract. E.g.: Simplify 5 7 Step : Fid the lowest Lowest commo deomiator is. Step : Write equivalet fractios with the 5 6 Step : Write both umerators over a sigle Step : Subtract the umerators. 9 7 5 6 Step 5: Reduce, if ecessary 9 is already reduced. Subtractig ratioal epressios is similar ecept that we have to determie the o-permissible value(s) for the epressio so we ca write the restrictios. E.g.: Simplify m m Step : Fid the lowest Lowest commo deomiator is. Step : Write both umerators over a sigle mm Use paretheses to show that you are subtractig all of the secod deomiator. 6

Step : Chage to additio ad chage the sigs of m m mm all the terms iside the paretheses. Step : Add like terms. Step 5: Reduce, if ecessary is already reduced. Step 6: Fid the o-permissible values. 0 Step 7: Write the aswer with ay restrictios ; 0 E.g.: Simplify 0 m 8 m m m Step : Fid the lowest Lowest commo deomiator is m. Step : Write both umerators over a sigle 0m 8 m Use paretheses to show m that you are subtractig all of the secod deomiator. Step : Chage to additio ad chage the sigs of 0m 8 m 0m8 m all the terms iside the paretheses. m m Step : Combie like terms. m 9 m Step 5: Reduce, if ecessary m 9 m m m Step 6: Fid the o-permissible values. m Step 7: Write the aswer with ay restrictios ; m E.g.: Simplify 6 Step : Factor both deomiators. The deomiators caot be factored.. Step : Fid the lowest Lowest commo deomiator is Step : Write both epressios with the commo 6 6 deomiator. Step : Write both umerators over a sigle 6 7

Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary 6 is already reduced. Step 8: Fid the o-permissible values., Step 9: Write the aswer with ay restrictios ;, For may ratioal epressios, the deomiators are NOT like, so we have to fid a commo deomiator before we ca subtract. We may have to factor the deomiators before we ca fid the E.g.: Simplify p p Step : Factor both deomiators. p p p p p p Step : Fid the lowest Lowest commo deomiator is p p Step : Write both epressios with the commo deomiator. p p p p p p p p p p Step : Write both umerators over a sigle p p p Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary p p p p p p p p p is already reduced. Step 8: Fid the o-permissible values. p Step 9: Write the aswer with ay restrictios p ; p p p. 8

E.g.: Simplify Step : Factor both deomiators. Step : Fid the lowest Lowest commo deomiator is Step : Write both epressios with the commo deomiator. Step : Write both umerators over a sigle Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary is already reduced. Step 8: Fid the o-permissible values. 0, Step 9: Write the aswer with ay restrictios ; 0,. E.g.: Simplify 0 6 5 0 Step : Factor both the umerators ad the deomiators ad cacel ay like factors. 9 5 5 7 5 0 7 6 0 0 0 Step : Fid the lowest Lowest commo deomiator is 7 0 Step : Write both epressios with the commo 0 0 deomiator. 7 7 0 7 0 7 7 0 0 7 7 0 Step : Write both umerators over a sigle 0 7 70.

Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary 0 70 70 70 is already reduced. Step 8: Fid the o-permissible values. 7,,5,0 Step 9: Write the aswer with ay restrictios ; 7,,5,0 7 0 E.g.: Simplify 6 Step : Factor both the umerators ad the deomiators ad cacel ay like factors. 6 Step : Fid the lowest Step : Write both epressios with the commo deomiator. Step : Write both umerators over a sigle Lowest commo deomiator is. Step 5: Epad the umerator, if ecessary. Step 6: Combie like terms. Step 7: Reduce, if ecessary 5 5 is already reduced. Step 8: Fid the o-permissible values.,, Step 9: Write the aswer with ay restrictios 5 ;,,, 0

E.g.: Fid the error i the simplificatio below ad write the correct simplificatio. 6 ; The error occurs i the subtractio of. The correct simplificatio is ; Do # s a, c, 5 a, c, 6 b, d, 7 b, 8, 9, CYU/Practisig pp. 9-50 tet i your homework booklet.

.5 Solvig Ratioal Equatios ( classes) Read Goal p. 5 tet Outcomes:. Defie ad give a eample of a ratioal equatio. p. 5. Solve ratioal equatios algebraically. pp. 5-57. Idetify the restrictio(s) o the variable(s) i a ratioal equatio. pp. 5-57. Idetify etraeous roots whe solvig a ratioal equatio. p. 5 5. Idetify iadmissible roots whe solvig a ratioal equatio. p. 5 Def : A ratioal equatio is a equatio that cotais oe or more ratioal epressios. E.g.: d d t t, 8, v v 0 0 We ca ofte solve ratioal equatios by multiplyig each term of the equatio by the LCD of all the deomiators i the equatio. We have to determie the restrictios o the variables i the ratioal equatios to idetify ay etraeous roots or iadmissible solutios. Def :A etraeous root is a solutio to the equatio that is ot permissible i the origial equatio. Def :A iadmissible solutio is a solutio to the equatio that is ot valid i the cotet of the problem. E.g.: Solve 6 t, t 0 t Step : Fid the lowest The LCD is t. Step : Multiply each term of the equatio by the 6 t t t t LCD. t Step : Simplify each term. There should be o t 8t fractios remaiig. Step : Rearrage to get a quadratic equatio. t 8t 0 t6 t 0 Step 5: Factor or use the quadratic formula. Step 6: Use ZPP to get the solutios. Step 7: Check for etraeous roots or iadmissible solutios. ZPP t 6 0 or t 0 t 6 or t The restrictio is t 0 so both ad 6 are solutios. Aother way to solve ratioal equatios is to combie epressios o each side of the equal side ad cross-multiplyig. E.g.: Solve 6 t, t 0 t

Step : Fid the lowest The LCD is t. Step : Write each term o the left with a 6 t t deomiator of t. t t t t t Step : Write both epressios o the left over the t t Step : Cross multiply. t 8t Step 5: Rearrage to get a quadratic equatio. t 8t 0 t6 t 0 Step 6: Factor or use the quadratic formula. Step 7: Use ZPP to get the solutios. Step 8: Check for etraeous roots or iadmissible solutios. ZPP t 6 0 or t 0 t 6 or t The restrictio is t 0 so both ad 6 are solutios. E.g.: Solve,,0 5 Step : Fid the lowest The LCD is 5. Step : Multiply each term of the equatio by the 5 LCD. 5 5 5 Step : Simplify each term. There should be o 5 5 fractios remaiig. Step : Epad ad combie like terms. 5 5 0 00 9 Step 5: Solve the equatio. 9 Step 6: Check for etraeous roots or The restrictios are,0 so is a solutio. iadmissible solutios. E.g.: Solve 0 5, 0, Step : Fid the lowest The LCD is. Step : Multiply each term of the equatio by the LCD. 0 5

Step : Simplify each term. There should be o fractios remaiig. 0 5 Step : Epad ad combie like terms. 0 8 5 5 Step 5: Solve the equatio. Step 6: Check for etraeous roots or The restrictios are 0, so is a etraeous iadmissible solutios. root. Therefore we write. 9 8 E.g.: Solve,,6 y y 6 y 9y 8 Step : Factor the deomiators. Step : Fid the lowest commo deomiator. Step : Multiply each term of the equatio by the LCD. Step : Simplify each term. There should be o fractios remaiig. Step 5: Epad ad combie like terms. Step 6: Solve the equatio. Step 7: Check for etraeous roots or iadmissible solutios. y y y6 y6 y 9y 8 y y 6 The LCD is y y 6. 9 8 y y 6 y y 6 y y 6 y y 6 y y 6 y y 9 6 8 9y5 y 8 5y 8 5y 60 y The restrictios are,6 so is a solutio. 5 5 E.g.: Solve,, 6 Step : Factor the deomiators. Step : Fid the lowest commo deomiator. 6 The LCD is.

Step : Multiply each term of the equatio by the LCD. Step : Simplify each term. There should be o fractios remaiig. Step 5: Epad ad combie like terms. Step 6: Factor or use the quadratic formula. Step 7: Use ZPP to get the solutios. Step 7: Check for etraeous roots or iadmissible solutios. 5 5 5 5 9 5 0 5 5 0 5 0 9 5 5 0 5 0 5 0 ZPP 0 or 5 0 5 or The restrictios are,, so we kow that oly 5 is a solutio. See I Summary, p. 57 See http://www.regetsprep.org/regets/math/algtrig/ate/ratioaleqpract.htm for practice questios. Do # s,, 5 c, d, 6 b, d, CYU/Practisig p. 58 tet i your homework booklet. Solvig Problems Usig Ratioal Equatios ( classes) It costs Mr. Williams $500 to take some of his seior high Eglish studets to Triity to see a play. O the day of the trip, 0 studets do t show up which meas that each studet must pay a etra $.50 to cover the cost of the trip. How may studets actually wet o the trip? Let be the umber of studets who actually wet to see the play. Let 0 be the umber of studets who were origially goig to see the play. The smaller cost per studet with 0 studets who were origially goig is The larger cost per studet with oly the studets who actually wet is 500. Sice the differece i cost is.50 we ca write 500. 0 500 500.50; 0,0 0 5

We multiply each term by 0 to get rid of the fractios. This gives 500 500 0 500 0 500.5 0 0 0 0.50 500 5000 500.5 5 5000.5 5.5 5 5000 0.5 5 5000 0.5.5.5.5 0 000 0 ZPP 50 0 0 50 or 0 So 0 studets actually wet to see the play. Do you see that 50 is a iadmissible solutio because it is a root of 500 500.50; 0,0 but does ot make sese i the cotet of this problem because the 0 umber of studets goig o the trip caot be egative? Do # 7 Practisig p. 60 tet i your homework booklet. E.g.: Chris purchased a motorboat ad used it to travel 6 km upstream ad the 6 km back dowstream i a total of hours. If the river s curret is costat at km/h, algebraically determie a equatio which models this situatio ad use it to fid the speed of the boat i still water. Let v be the speed of the boat i still water. Let v be the speed of the boat goig dowstream. Let v be the speed of the boat goig upstream. Upstream Dowstream d 6km d 6km Speed v Speed v 6 t t 6 v v Sice the total time is hrs, we ca write 6

6 6 v v We multiply each term by v v 6 6 v v 6 6 to get rid of the fractios. This gives v v v v v v v v v v 6v 6v v vv 6 0 v v 6 0 v v v v 0 ZPP v or v Sice speed is positive, we kow that v (iadmissible root), so the speed of the boat i still water is km/h. Do #, Practisig p. 59 tet i your homework booklet. E.g.: Pria, workig aloe, ca pait a room i hours. If Paul works aloe, he ca pait the same room i hours. How log would it take to pait the room if they work together? Pria Paul Both Time to Pait Room (hrs) Fractio of Room Paited i Hour Fractio of Room Paited i Hours The equatio that models this situatio is or Multiplyig all the terms by 6 ad solvig gives 7

6 6 6 6 5 6 6.hrs or hr, miutes 5 It will take. hrs to pait the room if they both work together. Do # 0 Practisig p. 59 tet i your homework booklet. E.g.: Pria, workig aloe, ca pait a room i hours. If Pria ad Paul work together, they ca pait the same room i. hours. How log would it take Paul to pait the room if he works aloe? Pria Paul Both Time to Pait Room (hrs) Fractio of Room Paited i Hour.. Fractio of Room Paited i. Hours.......... The equatio that models this situatio is. or. Multiplyig all the terms by ad solvig gives...9.6. 9.6.. 9.6.6 9.6.6.6.6 6 It will take 6 hrs for Paul to pait the room aloe. Do # s, 8, Practisig p. 59 tet i your homework booklet. 8

E.g.: Mrs Harum bought a bo of cheer-leader T-shirts for $50. She kept oe for herself ad oe for her daughter ad sold the rest for $560, makig a profit of $5 o each T-shirt. How may T-shirts were i the case? Let represet the umber of T-shirts i the bo. Profit/shirt = Reveue/shirt Cost/shirt so we ca write the equatio 560 50 5,,0 Multiplyig both sides by gives 560 50 5 5 560 50 5 0 560 50 900 5 0 560 50 900 0 5 0 900 0 5 0 900 0 0 0 5 900 5 0 00 8000 0 0 00 0 0 80 0 80 or 0 0 0 or 6 Sice 6 is a iadmissible root, the there were 0 shirts i the bo. Do # s 5, 8, Practisig p. 60 tet i your homework booklet. Do # s, d, b, d, 5 c, d, 6 a, d, 7, 9 a, c, 0 b, d,, Practisig p. 65-66 tet i your homework booklet. 9