Expressions and Equations

Slides:



Advertisements
Similar presentations
Copyright © Cengage Learning. All rights reserved. 6 Equations and Formulas.
Advertisements

ALGEBRA. PURPOSE Algebra is a building block that we can use to learn more advanced branches of mathematics such as statistics and calculus.
QUADRATICS EQUATIONS/EXPRESSIONS CONTAINING x2 TERMS.
Solving Linear Equations
Expressions and Equations
2.1 – Linear Equations in One Variable
Warm up Use the laws of exponents to simplify the following. Answer should be left in exponential form.
Solving Algebraic Equations
ALGEBRA, EQUATIONS AND FORMULAE. INTRODUCTION  Algebra essentially involves the substitution of letters for numbers in calculations, so that we can establish.
Solve the following: (8 + v)2 – 10 = 22
ALGEBRAIC EQUATIONS. EQUATIONS AND SOLUTIONS  A correct equation is like a balance scale.  In order to determine if a given value for a variable is.
MATHEMATICS Algebraic Manipulation – Multiplying and Dividing Terms.
Solving One and Two step Equations
Copyright © 2010 Pearson Education, Inc
Linear Algebra Achievement Standard 1.4.
Algebra 1 Final Exam Review – 5 days (2nd Semester)
Demonstrate Basic Algebra Skills
A. b. c By Jon Parkinson.
Methods and Solving Equations
ALGEBRA 1.
Algebra 1 Final Exam Review – 5 days (2nd Semester)
§ 2.8 Solving Linear Inequalities. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Linear Inequalities in One Variable A linear inequality in one.
Intermediate Tier - Algebra revision Contents : Collecting like terms Multiplying terms together Indices Expanding single brackets Expanding double.
Algebra Notes Algebra contains formulas, variables, expressions, equations, and inequalities. All of these things help us to solve problems.
Chapter 1 Review College Algebra Remember the phrase “Please Excuse My Dear Aunt Sally” or PEMDAS. ORDER OF OPERATIONS 1. Parentheses - ( ) or [ ] 2.
Solving Equations Medina1 Variables on Both Sides.
Section 2.2 More about Solving Equations. Objectives Use more than one property of equality to solve equations. Simplify expressions to solve equations.
2.1 The Addition Property of Equality
Chapter 1 Review Advanced Algebra 1.
SIMPLIFYING ALGEBRAIC EXPRESSIONS
12.1 Solving Two-Step Equations
By Kendal Agbanlog 6.1-Measurement Formulas and Monomials 6.2-Multiplying and Dividing Monomials 6.3-Adding and Subtracting Polynomials 6.4-Multiplying.
 Term- number or product of a number and a variable  Constant- term with no variable  Coefficient- number that multiplies to a variable  Like Terms-
Solving One and Two step Equations Sol A.1. Ex. 1) Solve r + 16 = -7 Think of this equation as a balance scale. Whatever you do to one side has to be.
Algebra.
Special Products a(x + y + z) = ax + ay + az (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy +y 2 (x – y) 2 = x 2 – 2xy +y 2 (x + y + z) 2 = x 2 + y.
Rules to Remember When solving an equation, the goal is to get the variable by itself. Addition and Subtraction are inverse operations. (opposites) Multiplication.
Algebra 1 Shelby Ferreira. Vocabulary Variable Coefficient Exponent Like terms Expression Equation.
Algebra Expressions Year 9.
Chapter 3: Real Numbers and the Coordinate Plane.
STROUD Worked examples and exercises are in the text 1 STROUD Worked examples and exercises are in the text Programme F2: Introduction to algebra PROGRAMME.
Transparency 1 Click the mouse button or press the Space Bar to display the answers.
Algebraic Expressions. Basic Definitions A term is a single item such as: An expression is a collection of terms d 5b -2c 3c2c3d2a 2a+3a3b-b4g-2g+g.
Solving two step Inequalities < < < > < > <
Algebra 1 Shelby Ferreira. Group Activity Half a number plus 5 is 11.What is the number? Explain your reasoning and create an equation that represents.
Section 6.2 Solving Linear Equations Math in Our World.
Algebra.
1.4 Solving Equations.
2. Algebraic manipulation
A1 Algebraic manipulation
KS4 Mathematics A2 Linear equations.
Algebra Skills Year 10.
LESSON 1.11 SOLVING EQUATIONS
Solving Two- Step Equations
Objective 3.6 solve multi-step inequalities.
ALGEBRA 1.
Solving Two- Step Equations
Introduction to Algebra
Solving Two- Step Equations
Objective Solve equations in one variable that contain more than one operation.
Using Algebra to Solve Equations
AM1 Algebraic Manipulation
Use of symbols Objectives:
Solving Two- Step Equations
Objective Solve equations in one variable that contain more than one operation.
KS4 Mathematics A2 Linear equations.
Solving Equations.
Presentation transcript:

Expressions and Equations ALGEBRA Expressions and Equations

Creating expressions Note: × and ÷ are not often used in Algebra i.e. 5 × x = 5x i.e. 8 ÷ x = 8 x Also a dot ‘.’ means multiply i.e. 2x . 2y = 2x × 2y Creating expressions - Using suitable symbols to express rules e.g. Write an expression for each of the following Let n = a number a) A number with 12 added to it n + 12 b) A number with 9 subtracted from it n - 9 c) A number multiplied by 2 n × 2 Best written as 2n d) A number divided by 6 n ÷ 6 Best written as n 6 As long as you explain what a symbol represents, any symbol can be used

e.g. John has x dollars. How much will he have if: a) He spends $35 x - 35 50 - 35 = $15 x + 28 50 + 28 = $78 b) He is given $28 c) He doubles his money 2 x 2 x 50 = $100 d) He spends half x 50 = $25 2 2 Once you have an expression, it can be used to calculate values if you know what the ‘variable’ (symbol) is worth. e.g. John has $50, use the expressions to calculate how much he will have in each situation:

Simplifying expressions by multiplying - ALL terms can be multiplied Rules: 1) Multiply all numbers in the expression 2) Place letters in alphabetical order behind product e.g. Simplify: a) p × 2q = 1 × 2 × p × q b) 2a × 5b = 2 × 5 × a × b = 2pq = 10ab c) 3 × 4x × 2y = 3 × 4 × 2 × x × y No number = 1 i.e. p = 1p = 24xy LIKE and UNLIKE TERms - LIKE terms are those with exactly the same letter or combination of letters LIKE terms: 2x, 3x, 31x UNLIKE terms: 2x, 3 4ab, 7ab 5p, 6q 2ab, 2ac e.g. Circle the LIKE terms in the following groups: a) 3a 5b 6a 2c b) 2xy 4x 12xy 3z 4yx While letters should be in order, terms are still LIKE if they are not.

Simplifying by adding/subtracting - We ALWAYS aim to simplify expressions from expanded to compact form - Only LIKE terms can be added or subtracted - When adding/subtracting just deal with the numbers in front of the letters e.g. Simplify these expanded expressions into compact form: a) a + a + a 1 1 1 = (1 + 1 + 1)a b) 5x + 6x + 2x = (5 + 6 + 2)x = 3a = 13x c) 3p + 7q + 2p + 5q = (3 + 2)p (+ 7 + 5)q d) 4a + 3b + 7a + b 1 = 5p + 12q = (4 + 7)a (+ 3 + 1)b = 11a + 4b - For expressions involving both addition and subtraction take note of signs e.g. Simplify the following expressions: a) 4x + 2y – 3x = (4 – 3)x + 2y b) 3a – 4b – 6a + 9b = x + 2y = (3 - 6)a (- 4 + 9)b = -3a + 5b c) -5d + 4a + 2 + 6d - 5a = (+ 4 - 5)a (- 5 + 6)d + 2 = -a + q + 2 If the number left in front of a letter is 1, it can be left out d) 2a – 5b – 6c = 2a – 5b – 6c

SUBSTITUTION - Involves replacing variables with numbers and calculating the answer - Remember the BEDMAS rules e.g. If a = 7, b = 3, and c = 5, calculate the following a) a + b + c = 7 + 3 + 5 b) b - a = 3 - 7 = 15 = -4 c) abc = 7 × 3 × 5 = 105 - Or replacing numbers into formulas e.g. The area of a triangle is given by the formula A = ½b × h Find the area of: Area = ½ × 12 × 5 5 m = 30 m2 12 m - Can substitute using mixed operations e.g. If a = 2, and b = 5, calculate the following a) 5b – 4a = 5 × 5 – 4 × 2 b) ab + 15 = 2 × 5 + 15 = 25 - 8 = 10 + 15 = 17 = 25

Because the top needs to be calculated first, brackets are implied - But always calculate brackets first! e.g. If a = 2, and b = 5, calculate the following a) 4(a + b) = 4(2 + 5) b) 2b – a 4 = (2 × 5 – 2) 4 = 4 × 7 = (10 – 2) 4 = 28 = 8 4 = 2 - Can involve powers too which are done after brackets e.g. If a = 2, and b = 5, calculate the following a) a2 = 22 b) (b – 2)2 = (5 – 2)2 = 4 = 32 = 9 c) 4a2 = 4 × 22 d) 2b2 – 3a2 = 2 × 52 - 3 × 22 = 4 × 4 = 2 × 25 - 3 × 4 = 16 = 50 - 12 = 38

POwers - Remember: 3 × 3 × 3 × 3 = 34 - Variables (letters) that are multiplied by themselves are treated the same way e.g. Simplify these expressions that are written in full a) r × r = r2 b) p × p × p × p × p = p5 - Sometimes there may be two or more variables e.g. Simplify a) a × a × b × b × b = a2b3 b) d × e × e × d × f = d2e2f Letters should still be written in alphabetical order! - With numbers and variables, multiply numbers and combine variables e.g. Simplify a) 4x × 6x = 4 × 6 × x × x b) 6b × 3b × 2 = 6 × 3 × 2 × b × b = 24x2 = 36b2 c) 2a × 5b × 3a × b 1 = 2 × 5 × 3 × 1 × a × a × b × b = 30a2b2

Multiplying using POwers - Does x2 × x3 = x × x × x × x × x ? YES - Therefore x2 × x3 = x5 - How do you get 2 3 = 5 ? + - When multiplying index (power) expressions with the same letter, ADD the powers. No number = 1 i.e. p = 1p1 e.g. Simplify a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1 = p12 = a6 - Remember to multiply any numbers in front of the variables first e.g. Simplify a) 2x3 × 3x4 = 2 × 3 x(3 + 4) b) 2a2 × 3a × 5a4 1 = 6 x7 = 2 × 3 × 5 a(2 + 1 + 4) = 30 a7

dividing using POwers - Does 6 = 1 ? 6 YES - Therefore x = 1 x - Does x5 = x × x × x × x × x ? x3 x × x × x YES = x × x × 1 × 1 × 1 - Therefore x5 = x2 x3 - How do you get 5 3 = 2 ? - - When dividing index (power) expressions with the same letter, SUBTRACT the powers. e.g. Simplify a) p5 ÷ p = p(5 - 1) b) x7 x4 = x(7 - 4) 1 = p4 = x3 - Remember to divide any numbers in front of the variables first e.g. Simplify a) 12x5 ÷ 6x4 = 12 ÷ 6 x(5 - 4) b) 5a7 15a2 ÷ 5 = 1 5 a(7 - 2) = 2 x = 1 5 a5 or a5 5 If the power remaining is 1, it can be left out

Expanding expressions - Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? YES 6 × 8 = 18 + 30 48 = 48 - The removal of the brackets is known as the distributive law and can also be applied to algebraic expressions - When expanding, simply multiply each term inside the bracket by the term directly in front e.g. Expand a) 6(x + y) = 6 × x + 6 × y b) -4(x – y) = -4 × x - -4 × y = 6x + 6y = -4x + 4y c) -4(x – 6) = -4 × x - -4 × 6 d) 7(3x – 2) = 7 × 3x - 7 × 2 = -4x + 24 = 21x - 14 e) x(2x + 3y) = x × 2x 1 1 + x × 3y f) -3x(2x – 5) = -3x × 2x - -3x × 5 = 2x2 + 3xy = -6x2 + 15x Don’t forget to watch for sign changes!

- If there is more than one set of brackets, expand them all then collect any like terms. e.g. Expand and simplify a) 2(4x + y) + 8(3x – 2y) = 2 × 4x + 2 × y + 8 × 3x - 8 × 2y = 8x + 2y + 24x - 16y = 32x - 14y b) -3(2a – 3b) – 4(5a + b) = -3 × 2a - -3 × 3b - 4 × 5a + -4 × 1b = -6a + 9b - 20a - 4b = -26a + 5b

factorising expressions - Factorising is the reverse of expanding - To factorise: 1) Look for a common factor to put outside the brackets 2) Inside brackets place numbers/letters needed to make up original terms You should always check your answer by expanding it e.g. Factorise a) 2x + 2y = 2( ) x + y b) 2a + 4b – 6c = 2( ) a + 2b - 6c - Always look for the highest common factor e.g. Factorise a) 6x - 15 = 3( ) 2x - 5 b) 30a + 20 = 10( ) 3a + 2 - Sometimes a ‘1’ will need to be left in the brackets e.g. Factorise a) 6x + 3 = 3( ) 2x + 1 b) 20b - 10 = 10( ) 2b - 1

- Letters can also be common factors e.g. Factorise a) cd - ce = c( ) d - e b) xyz + 2xy – 3yz = y( ) xz + 2x - 3z c) 4ad – 8a = 4 ( ) a d - 2 - Powers greater than 1 can also be common factors e.g. Factorise a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b

SOLVING EQUATIONS - When solving we need to isolate the unknown variable to find its value - To isolate we work backwards by undoing operations 1) To undo multiplication we use division e.g. Solve 3x = 18 ÷3 ÷3 x = 6 2) To undo addition we use subtraction e.g. Solve x + 2 = 6 -2 -2 x = 4 3) To undo subtraction we use addition e.g. Solve x - 8 = 11 +8 +8 x = 19 4) To undo division we use multiplication e.g. Solve x = 6 5 ×5 ×5 x = 30

Don’t forget the integer rules! - Terms containing the variable (x) should be placed on one side (often left) e.g. Solve a) 5x = 3x + 6 b) -6x = -2x + 12 -3x -3x +2x +2x 2x = 6 -4x = 12 Don’t forget the integer rules! ÷2 ÷2 ÷-4 ÷-4 x = 3 x = -3 Always line up equals signs and each line should contain the variable and one equals sign You should always check your answer by substituting into original equation Always look at the sign in front of the term/number to decide operation - Does 5×3 = 3×3 + 6 ? YES - Does -6×-3 = -2×-3 + 12 ? YES 15 = 9 + 6 18 = 6 + 12 - Numbers should be placed on the side opposite to the variables (often right) e.g. Solve a) 6x – 5 = 13 b) -3x + 10 = 31 +5 +5 -10 -10 6x = 18 -3x = 21 ÷6 ÷6 ÷-3 ÷-3 x = 3 x = -7

- Same rules apply for combined equations e.g. Solve a) 5x + 8 = 2x + 20 b) 4x - 12 = -2x + 24 -2x -2x +2x +2x 3x + 8 = 20 6x - 12 = 24 -8 -8 +12 +12 3x = 12 6x = 36 ÷3 ÷3 ÷6 ÷6 x = 4 x = 6 - Answers can also be negatives and/or fractions e.g. Solve a) 8x + 3 = -12x - 17 b) 5x + 2 = 3x + 1 +12x +12x -3x -3x 20x + 3 = -17 2x + 2 = 1 -3 -3 -2 -2 20x = -20 2x = -1 ÷20 ÷20 ÷2 ÷2 x = -1 x = -1 2 Make sure you don’t forget to leave the sign too! Answer can be written as a decimal but easiest to leave as a fraction

- Expand any brackets first e.g. Solve a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8 3x + 3 = 6 6x - 2 = x + 8 -3 -3 -x -x 3x = 3 5x - 2 = 8 ÷3 ÷3 +2 +2 x = 1 5x = 10 ÷5 ÷5 x = 2

Writing EQUATIONS and solving - Involves writing an equation and then solving e.g. Write an equation for the following information a) I think of a number, multiply it by 3 and then add 12. The result is 36. Let n = a number 3 n + 12 = 36 a) I think of a number, multiply it by 5 and then subtract 4. The result is the same as if 18 were added to the number Let n = a number 5 n - 4 = n + 18 e.g. Write an equation for the following information and solve a) A rectangular pool has a length 5m longer than its width. The perimeter of the pool is 58m. Find its width x + 5 x + 5 + x + x + 5 + x = 58 4x + 10 = 58 Draw a diagram -10 -10 x x 4x = 48 Let x = width ÷4 ÷4 x + 5 x = 12 Therefore width is 12 m

b) I think of a number and multiply it by 7 b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number? Let n = a number 7 n = n 4 + 15 -4n -4n 3n = 15 ÷3 ÷3 n = 5 Therefore the number is 5