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

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3-2 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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3-3 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.1 Definitions AlgebraAlgebra applies quantitative concepts to unknown quantities represented by symbols constantA constant is a term whose value does not change E.g. variableA variable is a term that represents a quantity that may have different values E.g. x, y, z expressionAn (algebraic) expression is a combination of constants and variables using arithmetic operations E.g.

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3-4 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.1 Definitions (cont…) termA term is part of an expression that is connected to another term by a + or a – sign E.g. In the expression 2x + 6y – 4z, the terms are 2x, 6y and –4z coefficientA coefficient is a factor by which the rest of a term is multiplied E.g. The term 9xy has a coefficient of 9 degreeThe degree of expression is the highest exponent of any variable in the expression E.g. the expression 3x 2 + 9x + 5 is a quadratic or second degree expression

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3-5 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.1 Definitions (cont…) equationAn equation is a statement that two expressions are equal linear equationA linear 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 imultaneous equationsTwo simultaneous equations must be solved at the same time to find the values of the variables that will satisfy both equations

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3-6 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.1 Definitions (cont…) formulaA formula is a rule or principle that is expressed in terms of algebraic symbols E.g. Formula to find the area of a circle subjectFormulas can be rewritten to make another variable the subject E.g.

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3-7 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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:

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3-8 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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:

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3-9 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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 :

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3-10 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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3-11 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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3-12 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.3 Solving linear equations (cont…) 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 = 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

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3-13 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.3 Solving linear equations (cont…) 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 = x 15x x = 12 15x + 4x =

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3-14 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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3-15 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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)

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3-16 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.4 Solving simultaneous linear equations (cont...) Step 2: Eliminate the variable that has the same coefficient by: subtracting equation (4) from equation (3) signs are the samesubtract equations –If the signs are the same, then subtract equations signs are differentadd equations –If the signs are different, then add 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

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3-17 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 3.4 Solving simultaneous linear equations (cont...) Substitute in 3 for y in equation (1) and solve for x. 3x + 4(3) = 33 3x + 12 = 33 3x = x = 21 Divide both sides by 3 Therefore: x = 7

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3-18 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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3-19 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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 Solution Let the amount charged for parts (in dollars) = x Then the amount charged for labour (in dollars) = 2x Therefore, the cost of parts = $x = $76 and the cost of labour = $2x = $2 ( 76 ) = $152

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3-20 Copyright 2010 McGraw-Hill Australia Pty Ltd PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e 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

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