3-1 copyright 2010 mcgraw-hill australia pty ltd powerpoint slides to accompany croucher,...

3-1Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e

Chapter 3

Algebra

Introductory Mathematics & Statistics


Learning Objectives

• Understand and use algebraic terms

• Manipulate algebraic expressions

• Solve simple linear equations (using transposition)

• Solve simultaneous linear equations

• Solve business problems using simple algebra


3.1 Definitions• AlgebraAlgebra applies quantitative concepts to unknown quantities

represented by symbols

• A constantconstant is a term whose value does not changeE.g.

• A variablevariable is a term that represents a quantity that may have different valuesE.g. x, y, z

• An (algebraic) expressionexpression is a combination of constants and variables using arithmetic operationsE.g.

,8

5,5.3,12

xy5z3x 23 y3x2

yx


3.1 Definitions (cont…)

• A term term is part of an expression that is connected to another term by a + or a – signE.g. In the expression 2x + 6y – 4z, the terms are 2x, 6y

and –4z

• A coefficientcoefficient is a factor by which the rest of a term is multipliedE.g. The term 9xy has a coefficient of 9

• The degreedegree of expression is the highest exponent of any variable in the expressionE.g. the expression 3x2 + 9x + 5 is a quadratic or second

degree expression



• An equationequation is a statement that two expressions are equal

• A linear equationlinear equation is one in which the largest value of the exponents is 1

• Solving the equationSolving the equation is where we find the value for the variable which makes the equation a true statement

• Two simultaneous equationsimultaneous equations must be solved at the same time to find the values of the variables that will satisfy both equations



• A formulaformula is a rule or principle that is expressed in terms of algebraic symbolsE.g. Formula to find the area of a circle

• Formulas can be rewritten to make another variable the subjectsubjectE.g.

2rA

Ar


3.2 Algebraic expressions

• Rule:If an expression contains like terms, these terms may be combined into a single term. Like terms are terms that differ only in their numerical coefficient. Constants may also be combined into a single constant

Example:

xx-x

x-x

xx

235

termsingleaformto

combined bemay 3and5

termslikeare3and5

z4x3x5ressionexptheSimplify


3.2 Algebraic expressions (cont…)

• Rule:When an expression is contained in brackets, each term within the brackets is multiplied by any coefficient outside the brackets. This is called expanding

Example:

286124232

bracketstheremoveto

1432

:expressiontheConsider

yxyx

yx


3.2 Algebraic expressions (cont…)

• Rule:

To multiply one expression by another, multiply each term of one expression by each term of the other expression. The resulting expression is said to be the product of the two expressions.

Example:

2xx6

2x4x3x6

1x221x2x31x22x3

1x2and2x3

sexpressiontwotheofproductThe

2

2


3.3 Solving linear equations• When solving an equation that involves only one

variable, follow these steps:

1. Place all the terms involving the variable on the left-hand side of the equation and the constant terms on the right-hand side

2. Collect the like terms, and collect the constant terms

3. Divide both sides of the equation by the coefficient of the variable


3.3 Solving linear equations (cont…)

• The left-hand side of the equation should now consist of the variable only. The right-hand side of the equation is the solution

• When terms are moved from one side of an equation to the other, they are said to be transposed (or transferred). This process is called transposition



• For example:

Solve 9x – 27 = 4x + 3 for x

1. Place like terms of the variable on the left side of the equation and the constant terms on the right side

9x – 4x = 3 + 27

2. Collect like terms and constant terms

5x = 30

3. Divide both sides of the equation by the coefficient of the variable (in this case 5).

x = 6



• Rule:Any term may be transposed from one side of an equation to the other. When the transposition is made, the sign of the term must change from its original sign. That is, a ‘+’ becomes a ‘-’ and a ‘-’ becomes a ‘+’.

Example:

15x - 20 = 12 - 4x

15x - 20 + 4x = 12

15x + 4x = 12 + 20


3.4 Solving simultaneous linear equations

• We consider the situation in which we have to find the values of two variables

• We must have two equations involving the variables

• Such equations are referred to as simultaneous equations

• In this case, there will usually be unique values of the variables that will satisfy both equations


3.4 Solving simultaneous linear equations (cont...)

• Solve for x and y

3x + 4y = 33 (1)2x – 3y = 5 (2)

• Step 1:

Make the coefficient of either of the variables in one equation equal to its coefficient in the other equation. Multiply both sides of equation (1) by 2 and equation (2) by 3.

6x + 8y = 66 (3)

6x – 9y = 15 (4)



• Step 2:

Eliminate the variable that has the same coefficient by: subtracting equation (4) from equation (3)

– If the signs are the samesigns are the same, then subtract equationssubtract equations

– If the signs are differentsigns are different, then add equationsadd equations

6x + 8y = 66 (3)

minus: 6x – 9y = 15 (4)equals 8y – (– 9y) = 51

17y = 51

Divide both sides by 17 to find y

Therefore:

y = 3



• Substitute in 3 for y in equation (1) and solve for x.

3x + 4(3) = 33

3x + 12 = 33

3x =

33 - 12 3x

= 21

Divide both sides by 3

Therefore: x = 7


3.5 Applications

• There are many problems in which the value of some unknown quantity is to be found

• In many cases, we can represent the unknown quantity by some variable name and construct an equation involving that variable

• There is a vast array of such problems, and the best way to describe them is to look at different examples and then determine how they are solved


3.5 Applications (cont…)Example

Tran left his car for service and received a combined bill (for parts and labour) of $228. Given that the labour costs twice as much as the parts, find the amount that parts and labour each cost

SolutionLet the amount charged for parts (in dollars) = xThen the amount charged for labour (in dollars) = 2x

Therefore, the cost of parts = $x = $76 and the cost of labour = $2x = $2 ( 76 ) = $152

76x3

228

3

x3

228x3

228$x2x


Summary

• We have looked at different algebraic definitions

• We have also looked at simplifying algebraic expressions

• We examined solving simple linear equations, which is solving for one variable

• We also solved simultaneous linear equations, which is solving equations with two variables

• Lastly we had a brief look at the applications of algebra

3-1 copyright 2010 mcgraw-hill australia pty ltd powerpoint slides to accompany croucher,...

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