Equations and inequalities: Solving linear equations

Similar documents
Functions and graphs: The parabola (Grade 10) *

Quadratic Functions and Graphs *

Factorising Cubic Polynomials - Grade 12 *

Graphing Linear Equations and Inequalities: Proficiency Exam

Functions and graphs - Grade 10 *

FHSST: Grade 10 Maths. Collection Editor: Free High School Science Texts Project

Algebra II A Guided Notes

Trigonometry: Graphs of trig functions (Grade 10) *

Solving Linear Equations

Lesson 7: Watch Your Step!

Solving Linear Equations (in one variable)

NCERT solution for Integers-2

5.1 Simplifying Rational Expressions

Exponential Functions and Graphs - Grade 11 *

P1 Chapter 3 :: Equations and Inequalities

Solving Linear and Rational Inequalities Algebraically. Definition 22.1 Two inequalities are equivalent if they have the same solution set.

Algebra Revision Guide

Chapter 4: Radicals and Complex Numbers

Equations and inequalities

Exponential and Logarithmic Equations

Unit 6 Study Guide: Equations. Section 6-1: One-Step Equations with Adding & Subtracting

Linear And Exponential Algebra Lesson #1

Even and odd functions

Core 1 Inequalities and indices Section 1: Errors and inequalities

Rational Numbers CHAPTER. 1.1 Introduction

Core 1 Basic Algebra. Section 1: Expressions and equations

Polynomials. Eve Rawley, (EveR) Anne Gloag, (AnneG) Andrew Gloag, (AndrewG)

Lesson 3: Solving Equations A Balancing Act

A-LEVEL MATHS Bridging Work 2017

Algebra III and Trigonometry Summer Assignment

Representing chemical change: State symbols *

Roots of quadratic equations

A. Incorrect! This inequality is a disjunction and has a solution set shaded outside the boundary points.

Domain and range of exponential and logarithmic function *

Physical and Chemical change: Conservation of matter *

OCR Mathematics Advanced Subsidiary GCE Core 1 (4721) January 2012

BASIC ALGEBRA ALGEBRA 1. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Basic algebra 1/ 17 Adrian Jannetta

Roots of quadratic equations

Algebra Year 10. Language

OPTIONAL: Watch the Flash version of the video for Section 6.1: Rational Expressions (19:09).

Function Operations and Composition of Functions. Unit 1 Lesson 6

Edexcel New GCE A Level Maths workbook Solving Linear and Quadratic Simultaneous Equations.

Chapter Usual types of questions Tips What can go ugly. and, common denominator will be

CONTENTS CHECK LIST ACCURACY FRACTIONS INDICES SURDS RATIONALISING THE DENOMINATOR SUBSTITUTION

ALGEBRA 1. Interactive Notebook Chapter 2: Linear Equations

Intermediate Algebra

7.3 Adding and Subtracting Rational Expressions


UNIT 5 VOCABULARY: POLYNOMIALS

IES Parque Lineal - 2º ESO

Equations and Solutions

Polar Form of Complex Numbers

Factorisation CHAPTER Introduction

Solving Polynomial and Rational Inequalities Algebraically. Approximating Solutions to Inequalities Graphically

3.1 Inequalities - Graphing and Solving

Numerical and Algebraic Fractions

Displacement. OpenStax College. This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License

Mathematics Revision Guide. Algebra. Grade C B

EQUATION OF A CIRCLE (CENTRE A, B)

Self-Directed Course: Transitional Math Module 4: Algebra

The Periodic Table - Grade 10 [CAPS] *

Chapter 2A - Solving Equations

Name Class Date. t = = 10m. n + 19 = = 2f + 9

Motion in two dimensions: vertical projectile motion *

Complex fraction: - a fraction which has rational expressions in the numerator and/or denominator

GRADUATE RECORD EXAMINATIONS. Math Review. Chapter 2: Algebra

Algebra Year 9. Language

Learning Module 1 - Basic Algebra Review (Appendix A)

Solving Absolute Value Equations and Inequalities

Rational and Radical Relationships

AS Mathematics AS Further Mathematics Transition Booklet. Name: Bridging the gap to A Level!

Expanding brackets and factorising

Intermediate Algebra Textbook for Skyline College

Algebra 2 Prep. Name Period

Linear Equations in One Variable *

Westside. Algebra 2 PreAP

Factorizing Algebraic Expressions

5.6 Solving Equations Using Both the Addition and Multiplication Properties of Equality

Westside Algebra 2 PreAP

Factoring and Algebraic Fractions

Project IV Fourier Series

CfE Higher Mathematics Course Materials Topic 4: Polynomials and quadratics

A summary of factoring methods

DIVIDING BY ZERO. Rational Expressions and Equations. Note Package. Name: 1: Simplifying Rational Expressions 2: Multiplying and Dividing

OCR Mathematics Advanced Subsidiary GCE Core 4 (4724) June 2010

SIXTH FORM MATHEMATICS A LEVEL INDUCTION BOOKLET SEPTEMBER Name:

Solutions and solubility - Grade 11

A Level Maths. Induction Booklet CONTENTS

Section 1.1 Task List

Preparing for A-Level Mathematics Summer 2017

This is Solving Linear Systems, chapter 3 from the book Advanced Algebra (index.html) (v. 1.0).

Unit 3. POLYNOMIALS AND ALGEBRAIC FRACTIONS.

Equations, Inequalities, and Problem Solving

Zeros of Polynomial Functions

Maths Scheme of Work. Class: Year 10. Term: autumn 1: 32 lessons (24 hours) Number of lessons

Northwood High School Algebra 2/Honors Algebra 2 Summer Review Packet

A Level Maths summer preparation work

Chapter 1.6. Perform Operations with Complex Numbers

Quadratics. SPTA Mathematics Higher Notes

DIRECTED NUMBERS ADDING AND SUBTRACTING DIRECTED NUMBERS

Transcription:

Connexions module: m39780 1 Equations and inequalities: Solving linear equations Free High School Science Texts Project This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License 1 Strategy for Solving Equations This chapter is all about solving dierent types of equations for one or two variables. In general, we want to get the unknown variable alone on the left hand side of the equation with all the constants on the right hand side of the equation. For example, in the equation x 1 = 0, we want to be able to write the equation as x = 1. As we saw in review of past work 1 (section on rearranging equations), an equation is like a set of weighing scales that must always be balanced. When we solve equations, we need to keep in mind that what is done to one side must be done to the other. 1.1 Method: Rearranging Equations You can add, subtract, multiply or divide both sides of an equation by any number you want, as long as you always do it to both sides. For example, in the equation x+5 1 = 6, we want to get x alone on the left hand side of the equation. This means we need to subtract 5 and add 1 on the left hand side. However, because we need to keep the equation balanced, we also need to subtract 5 and add 1 on the right hand side. x + 5 1 = 6 x + 5 5 1 + 1 = 6 5 + 1 x + 0 + 0 = 11 + 1 x = 10 (1) In another example, 3x = 8, we must divide by and multiply by 3 on the left hand side in order to get x alone. However, in order to keep the equation balanced, we must also divide by and multiply by 3 on the Version 1.1: Aug, 011 10:00 am GMT-5 http://creativecommons.org/licenses/by/3.0/ 1 "Review of Past Work" <http://cnx.org/content/m31330/latest/>

Connexions module: m39780 right hand side. 3 x = 8 3 x 3 = 8 3 3 3 x = 8 3 1 1 x = 1 x = 1 These are the basic rules to apply when simplifying any equation. In most cases, these rules have to be applied more than once, before we have the unknown variable on the left hand side of the equation. () tip: The following must also be kept in mind: 1.Division by 0 is undened..if x y = 0, then x = 0 and y 0, because division by 0 is undened. We are now ready to solve some equations! 1.1.1 Investigation : Strategy for Solving Equations In the following, identify what is wrong. x 8 = 3 (x ) (x ) = 3 (x ) (x ) (x ) = 3(x ) (x ) = 3 (3) Solving Linear Equations The simplest equation to solve is a linear equation. A linear equation is an equation where the power of the variable(letter, e.g. x) is 1(one). The following are examples of linear equations. x + = 1 x = 3 x 6 = 7x + In this section, we will learn how to nd the value of the variable that makes both sides of the linear equation true. For example, what value of x makes both sides of the very simple equation, x + 1 = 1 true. Since the denition of a linear equation is that if the variable has a highest power of one (1), there is at most one solution or root for the equation. This section relies on all the methods we have already discussed: multiplying out expressions, grouping terms and factorisation. Make sure that you are comfortable with these methods, before trying out the work in the rest of this chapter. () x + = 1 x = 1 (liketermstogether) x = 1 (simplifiedasmuchapossible) (5)

Connexions module: m39780 3 Now we see that x = 1. This means if we divide both sides by, we will get: x = 1 If we substitute x = 1, back into the original equation, we get: (6) LHS = x + and = ( 1 ) + = 1 + = 1 RHS = 1 That is all that there is to solving linear equations. tip: Solving Equations When you have found the solution to an equation, substitute the solution into the original equation, to check your answer..1 Method: Solving Equations The general steps to solve equations are: 1. Expand (Remove) all brackets.. "Move" all terms with the variable to the left hand side of the equation, and all constant terms (the numbers) to the right hand side of the equals sign. Bearing in mind that the sign of the terms will change from (+) to ( ) or vice versa, as they "cross over" the equals sign. 3. Group all like terms together and simplify as much as possible.. Factorise if necessary. 5. Find the solution. 6. Substitute solution into original equation to check answer. Khan academy video on equations - 1 This media object is a Flash object. Please view or download it at <http://www.youtube.com/v/f15za0phsek&rel=0> (7) Figure 1 Exercise 1: Solving Linear Equations (Solution on p. 5.) Solve for x: x = Exercise : Solving Linear Equations (Solution on p. 5.) Solve for x: (x 9) x = 6x Exercise 3: Solving Linear Equations (Solution on p. 6.) x Solve for x: = Exercise : Solving Linear Equations (Solution on p. 6.) Solve for x: 3 x 6 = 7x +

Connexions module: m39780.1.1 Solving Linear Equations 1. Solve for y: y 3 = 7 Click here for the solution. Solve for w: 3w = 0 Click here for the solution 3 3. Solve for z: z = 16 Click here for the solution. Solve for t: 1t + 0 = 1 Click here for the solution 5 5. Solve for x: 7 + 5x = 6 Click here for the solution 6 6. Solve for y: 55 = 5y + 3 Click here for the solution7 7. Solve for z: 5z = 3z + 5 Click here for the solution 8 8. Solve for a: 3a 1 = 6 + a Click here for the solution 9 9. Solve for b: 1 6b + 3b = b 6 Click here for the solution 10 10. Solve for c: 6c + 3c = 5 (c 3) Click here for the solution 11 11. Solve for p: 18 p = p + 9 Click here for the solution 1 1. Solve for q: q = 16 Click here for the solution13 13. Solve for q: 1 = q Click here for the solution1 1. Solve for r: ( 16 r) = 13r 1 Click here for the solution 15 15. Solve for d: 6d + d = + d + 8 Click here for the solution 16 16. Solve for f: 3f 10 = 10 Click here for the solution. Solve for v: 3v + 16 = v 10 Click here for the solution 18 18. Solve for k: 10k + 5 + 0 = k + 3k + 80 Click here for the solution 19 19. Solve for j: 8 (j ) = 5 (j ) Click here for the solution 0 0. Solve for m: 6 = 6 (m + 7) + 5m Click here for the solution 1 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcr> 3 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcr> See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcr> 5 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcr> 6 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcr> 7 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcn> 8 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcn> 9 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcn> 10 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcn> 11 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcq> 1 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcq> 13 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcq> 1 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcq> 15 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcq> 16 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu> See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu> 18 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu> 19 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu> 0 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu> 1 See the le at <http://cnx.org/content/m39780/latest/http://www.fhsst.org/lcu>

Connexions module: m39780 5 Solutions to Exercises in this Module Step 1. We are given x = and are required to solve for x. Step. Since there are no brackets, we can start with grouping like terms and then simplifying. x = x = (moveallconstantterms (numbers) totherhs (righthandside)) x = 0 (groupliketermstogether) x = 0 (simplifygroupedterms) x = 0 x = 0 (8) Step. Substitute solution into original equation: 0 = (9) Since both sides are equal, the answer is correct. Step 5. The solution of x = is x = 0. = (10) Step 1. We are given (x 9) x = 6x and are required to solve for x. Step. We start with expanding the brackets, then grouping like terms and then simplifying. andallconstanttermstotherhsofthe = (x 9) x = 6x 8x 36 x = 6x (expandthebrackets) 8x x + 6x = + 36 movealltermswithxtothelhs (8x x + 6x) = ( + 36) (groupliketermstogether) 10x = 0 (simplifygroupedterms) 10 10 x = 0 10 (dividebothsidesby10) x = (11) Step. Substitute solution into original equation: ( () 9) () = 6 () (8 9) 16 = ( 1) 16 = 0 16 = 0 0 = 0 (1) Since both sides are equal to 0, the answer is correct. Step 5. The solution of (x 9) x = 6x is x =.

Connexions module: m39780 6 Step 1. We are given x = and are required to solve for x. Step. Since there is a denominator of (3x + 1), we can start by multiplying both sides of the equation by (3x + 1). But because division by 0 is not permissible, there is a restriction on a value for x. (x 1 3 ) andallconstantterms (numbers) totherhs. x = ( x) = (3x + 1) x = 6x + (remove/expandbrackets) x 6x = movealltermscontainingxtothelhs 7x = 0 (simplifygroupedterms) x = 0 ( 7) theref ore x = 0 zerodividedbyanynumberis0 (13) Step. Substitute solution into original equation: (0) 3(0)+1 = 1 = Since both sides are equal to, the answer is correct. x Step 5. The solution of = is x = 0. Step 1. We are given 3x 6 = 7x + and are required to solve for x. Step. We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying. 3 x 6 = 7x + andallconstanttermstotherhsofthe = Step. Substitute solution into original equation: x 18 = 1x + 6 (eachtermismultipliedby3) x 1x = 6 + 18 (movealltermswithxtothelhs x = (simplifygroupedterms) x = (dividebothsidesby ) x = (1) (15) 3 ( 8) 6 = 7 + () 6 = 7 ( ) + ( 3) 168 6 = + 3 10 = ( 168)+3 13 13 = Since both sides are equal to 13, the answer is correct. (16)

Connexions module: m39780 7 Step 5. The solution of 3x 6 = 7x + is, x =.