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ALGEBRA

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PURPOSE Algebra is a building block that we can use to learn more advanced branches of mathematics such as statistics and calculus.

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ANALYSIS TOOL It allows us to understand and critically evaluate the mathematics done by others, such as reporters, political candidates, insurance salesmen, bank loan officers, etc.

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APPLICATIONS IN EVERY DAY LIFE General Life Circumstances We use algebra in general life without even realising it. E.g. If we are traveling we need to know the distance we are going to travel. Algebra helps us in decision making; for example, it allows us to compare mobile phone companies and to decide which company offers better value for money. Understanding algebra helps us with data entry, understanding interest rates when we are investing, writing assignments, etc.

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Science and Careers Algebra is a prerequisite for advanced training in science fields such as chemistry, physics and mathematics. It is a key to opportunities which allows us to understand letters and symbols. It allows us to chose whether to take a detour in our careers or not.

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ALGEBRAIC NOTATION

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The most common notations and symbols used in algebra are the following:

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+ “plus”, denotes adding or the sum of… e.g. 5 + 6

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- “minus”, denotes subtracting or removing e.g. 10 - 4

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x/* “times”, denotes multiplications, an asterix can sometimes be used, meaning the same thing. e.g. 13 x 10 / 13 * 10

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NB The most common variables used are x and y Variables are the ones which are substituted by numbers, e.g. x can be substituted by 3, or y can be substituted by 9.

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Expression An expression shows a mathematical relationship whereby there is no solution

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E.g. 5x + 6 if x = 2, then x is substituted by 2 in the expression 5(2) + 6 = 10 + 6 = 16

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If x = 10, then 5(10) + 6 = 50 + 6 = 56

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Equation An equation looks like this: x + 3 = 5; The only difference between an equation and an expression is the equal sign (=)

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Formula A formula is a special equation; it shows the relationship between two variables. e.g. V= hwl Where V is the volume h is the height w is the width l is the length

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NB x + 3 = 5 is not a formula, it has only one variable x.

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Identity Identity is the relationship which is true. This means that whatever the number value may be, the answer stays the same. e.g. algebraically, this occurs if an equation is satisfied for all values of the involved variables.

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Examples

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Given a rectangle x+2 x Find the perimeter

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Solution: Write down an expression for perimeter. The perimeter is the distance right around the shape: x + x + (x + 2) + (x + 2) 2x + 2x + 4 4x + 4

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If x is 3, find the exact value of the perimeter. 4x + 4 is the expression of perimeter x is now equal to 3; substitute 3 in the expression 4(3) + 4 = 12 + 4 = 16 units

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MANIPULATING ALGEBRAIC EXPRESSIONS

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Simplify by collecting like terms Consider: a) 3a + 6b + 3a – 7b + a Solution: We have two terms a and b. Let’s group the a terms and the b terms together: 3a + 3a + a + 6b – 7b = 7a - b

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Consider: b) 2xzy – 8zy + 4xzy + 2y Solution: We now have three different terms: xyz, zy and y Let’s bring the like terms together: 2xzy + 4xzy - 8zy + 2y = 6xzy – 8zy + 2y

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Multiply out a single bracket Consider a) 2(5x – 3y) Solution: The number 2 outside the bracket multiplies everything inside the bracket. 2(5x – 3y) = 10x – 6y

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Consider b) -3(7z – 4y ) Solution: The number -3 will multiply everything inside the bracket. NB When a negative number multiplies a negative number a positive result is obtained and when a negative number and a positive number multiply a negative result is obtained

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-3(7z – 4y) = -21z + 12y ( we can rearrange this answer to make it look nice, but nothing changes) = 12y – 21z

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Factorise a single bracket by taking out a common factor consider: a) (2xy – 4xyz) Solution: We need to get factors which get into both 2xy and 4 xyz and factor them outside the single bracket

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These factors are 2, x and y 2xy(1 – 2z) ( divide the factors into the terms in the bracket) NB If you multiply the factor outside the bracket you will go back to the original expression

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Expand two brackets Consider: a) (x + y) (x + y) expand. Solution: each and every term will multiply the other. We start from the left to the right, then we group the terms.

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Factorise quadratics into brackets

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Factorise quadratics using the difference of two squares

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Simplify algebraic expressions by cancelling, adding, subtracting and multiplying

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To add or subtract algebraic fractions Do the following: 1. Factorise denominators ( if necessary) 2. Find the Lowest Common Factors of the denominators 3. Express each fraction in terms of this Lowest Common Factor and simplify.

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Use index laws, including fraction, zero and negative powers and powers raised to another power.

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LINEAR EQUATIONS

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A linear equation is an algebraic equation in which term is either a constant or a product of a constant and a single variable ( the first power) A linear equation can have one or more variables

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Examples of linear equations a) y = 3x – 2 b) y – 2 = 3(x + 1) c) 5x = 6

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Consider c) x + 2 = 2x – 4 Solution: Put like terms on one side 2 + 4 = 2x – x 6 = x

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Consider e) -2(x + 2) = 3 – x Solution: Simplify the brackets ( multiply by -2 in the bracket) -2x – 4 = 3 – x -2x + x = 3 + 4 group like terms -x = 7 multiply everything by -1 to get x x = -7

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2

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NB So in dealing with linear fractions, the following steps need to be followed. 1. Remove the denominator by multiplying every term in the equation by the Common Factor 2. Rearrange the equation 3. Put like terms to the same side 4. Divide both sides by the coefficient of x

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SIMULTANEOUS EQUATIONS

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Solving the equations by elimination Consider a) Solve for x and y 3x – 2y = 27 2x + y = 4 Label the equation 1 and 2 3x – 2y = 27 (1) 2x + y = 4 (2) Multiply equation (1) by 2 and equation (2) by 3 to make the coefficient of x the same

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NB After finding the first variable, you can substitute the variable in any of the equations to find the second variable.

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Solving simultaneous equations by substitution Consider b) Solve for x and y x + y = 45 (1) label equations 1 and 2 3x – 2y = 0 (2) Make x or y the subject of the formula in either of the equations. Replace x with 45 – y in equation 2 3(45 – y) – 2y = 0 Simplify the brackets

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135 – 3y – 2y = 0 Rearrange like terms 135 – 5y = 0 135 = 5y Divide both sides by 5 to get y 27 = y

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Substitute y for 27 in any of the two equations or use the one in which x was made the subject of the formula x = 45 – y x = 45 – 27 x = 18 x = 18; y = 27

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Consider c) The graph of y = 2x + 1 and y = 4 – 3x. Drawn on the same axes, find the coordinates of A where the graphs cross each other 0 A y=4-3x y=2x+1 1 X Y 4

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Solution: The two graphs are equal at the point of intersection Thus y = 2x + 1 = y = 4 – 3x y = 2x + 1 ( 1) Label the graphs y = 4 – 3x (2) 2x + 1 = 4 – 3x Since the two equations are equal to each other

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Word problem involving a simultaneous equation Consider d) The combined cost of a television and a DVD player is £ 1,460. The television costs £ 330 more than the DVD player (i) Use two equations in x and y to represent the situation (ii) Hence, find the cost of the television and the cost of the DVD player

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Solution: Let the cost of the television be x and the cost of the DVD player be y. (i) Thus x + y = 1460 (1) x = y + 330 ( since the TV costs £ 330 more than the DVD player, if we add £330 to the DVD, they become equal )

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x + y = 1460 (1) Label equations x = y + 330 (2) Replace x with y+330 in equation (1) y + 330 = y = 1460 rearrange 2y = 1460 – 330 2y = 1130divide by 2 to get y y = 565 From x = y + 330 x = 565 + 330 x = 895 The television costs £ 895 and the DVD player costs £ 565

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QUADRATIC EQUATIONS

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Solving quadratic equations by factorisation

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Solving quadratic equations by completing the square

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Solving a quadratic equation using a quadratic formula

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