Algebra Student Signature: Parent/Carer signature:

Preparing yourself for Mathematics at Guilsborough Algebra Student Signature: Parent/Carer signature:

Dear student, This booklet contains work which you need to complete over the summer. It is designed to help you develop key skills in mathematics which you need at secondary school. The first part of the booklet needs to be completed at home. The second part will be completed in lessons at school in September. You will need to bring the completed booklet to school on your first day. We look forward to seeing you in September Guilsborough Academy Mathematics Department

How does it work? Your Turn Algebra Basics 1 Phrases as algebraic expressions Write these phrases as algebraic expressions (let the number be n ) a The sum of a number and 7: Phrases as Algebraic Expressions.../.../0... ALGEBRA BASICS * ALGEBRA BASICS * n + 7 b The difference between 9 and a number: c The sum of 6 times a number and 1: d The product (#) of a number and 4: e The quotient (') of two more than a number and 3: f The difference between a number squared and 6: g The product of a number minus 5 and : h 8 less than twice a number: i 10 added to a number halved: j A number multiplied by 5 more than itself: 10 H 1 Algebra Basics SERIES TOPIC Mathletics Passport 3P Learning

How does it work? Algebra Basics Addition and subtraction + & If the variable parts are exactly the same, the terms are called like terms. Like terms: x Like terms - x 3b b y - 5y Like terms Like terms Not Like terms: a b p p x Not like terms Not like terms Not like terms - 5y Only like terms can be added or subtracted. Simplify a+ a a+ a Variable parts are the same (like terms) ` a+ a = 3a Simplify 8x- 3x 8x- 3x like terms ` 8x- 3x = 5x Simplify 3d+ 4d+ 6c d d c 3 4 6 + + like terms 3d+ 4d+ 6c = 7d+ 6c This cannot be simplified any further Why don t we add or subtract unlike terms? Good Question! Let s look at a problem the last example could represent. At a picnic for pets, each dog gets 7 treats and each cat gets 6 treats. Number of treats needed is: (7 treats # number of dogs) + (6 treats # number of cats) = (7 # d) + (6# c) the number of dogs the number of cats Simplified: = 7d+ 6c d and c represent two different animals so it does not make sense to add them together. Therefore 7d+ 6c is the simplest expression for this problem. 1 H 1 Algebra Basics SERIES TOPIC Mathletics Passport 3P Learning

How does it work? Expanding and Factorising Expanding Grouping symbols such as [brackets], {braces} and (parentheses), can be removed without changing the value of the expression by expanding. a^b+ ch = a # b and a # + c = ab + ac Every term inside the parentheses is multiplied by the term in front of the parentheses. Expand ^3x + 5h # ^3x+ 5h = ^3x+ 5h # every term inside the parentheses = # 3x and # + 5 = 6x and + 10 = 6x + 10 a^b+ ch = a # b and a #- c = ab -ac Expand 3a^a- 4h # 3aa ^ - 4h = 3aa ^ - 4h 3a # every term inside the parentheses = 3a # a and 3a #-4 = 3a and -1a = 3a -1a Now that you have seen how it works, it's time to learn the mathematical name we give this is method: The Distributive Law l l a^b+ ch = a # b and a # + c = ab + ac a^b- ch = a # b and a #- c = ab -ac I SERIES 3 TOPIC Expanding and Factorising Mathletics Passport 3P Learning

How does it work? Your Turn Expanding and Factorising Expanding EXPANDING * EXPANDING * EXPANDING * 1 Expand: a ^a + 7h b 9^b -3h.../.../0... c 6c^3d+ 1h d 4d^3- ch e 3x^6+ 4yh f 3m^p- qh g 1 ^ 6 m - 14 h h ab^3c+ dh i 4^-3-9xh j q -p`- j 4 I 3 Expanding and Factorising SERIES TOPIC Mathletics Passport 3P Learning

How does it work? Expanding and Factorising More expanding Why limit yourself to parentheses with only two terms? The Distributive Law works for parentheses with more. Every term inside the parentheses is multiplied by the term in front. Expand 4^m+ 3n-h # 4 ^ m+ 3n- h = 4 ^ m+ 3n- h 4 # every term inside the parentheses = 4 # m and 4 # + 3n and 4 # - = 8m and + 1n and - 8 = 8m+ 1n- 8 Take care with the multiplications when there is a negative term out the front. Expand -aa ^ - b+ 3c+ h # -aa ^ - b+ 3c+ h =-aa ^ - b+ 3c+ h a # every term inside the parentheses =-a # a and -a #-b and - a # + 3c and - a # + =- a and + ab and -3ac and - a =- a + ab-3ac- a The basic index laws are often used when expanding expressions. Expand p ^p- 3pq + 5qh # p ^p- 3pq + 5qh = p ^p- 3pq+ 5qh p # every term inside the parentheses = p # p and p #- 3pq and p # + 5q + 1 + 1 = p and - 3p q and + 5p q Remember: m n m+ n # = a a a 3 3 = p - 3p q+ 5p q 6 I 3 Expanding and Factorising SERIES TOPIC Mathletics Passport 3P Learning

How does it work? Your Turn Expanding and Factorising More expanding 1 Expand: a 3^a+ b+ h b 4^x-y-5h EXPANDING * MORE EXPANDING * MORE.../.../0... c 3p^p+ q + 4h d - de ^ + f + 6h e x^4x+ 3y- 3+ zh f -ab ^ - c+ d-5h Expand: (psst: remember the multiplication rule for indices) a nn ( +3n) 3 b xy^x - y h c - ab^ab + abh d p^p - 4pq + 5h Expanding and Factorising Mathletics Passport 3P Learning I SERIES 3 TOPIC 7

How does it work? Expanding and Factorising Expanding and simplifying Always simplify the expression after expanding where possible. Simplify by collecting like terms after the expansion of any parentheses. Expand and simplify: 3^7m-6h-16m # 37 ^ m-6h- 16m = 3^7m-6h-16m = 3# 7m and3# -6 and - 16m = 1m-18-16m Like terms = 5m -18 3 # every term inside the parentheses Combine the like terms For expressions with multiple parentheses, expand each separately then look to simplify. Expand and simplify: 5^a+ 4h-4^a-3h # # 5 ^ a+ 4h-4( a- 3) = 5^a+ 4) -4( a- 3h 5 # a and 5 # + 4-4 # a and -4 #-3 Like terms = 10a+ 0-4a+ 1 Like terms = 10a- 4a+ 0+ 1 = 6a + 3 Expand each grouping separately Identify the like terms Group the like terms Simplify Be careful to apply the index laws correctly when expanding expressions with multiple variables. Expand and simplify: xy^5x+ yh-xy # xy^5x+ yh- xy= xy^5x+ yh- xy xy # every term inside the parentheses = xy # 5x and xy # + y and - xy 1+ 1 1+ 1 = 5x y and xy and -xy = 5xy+ xy - xy Like terms = 5xy- xy+ xy = 3xy+ xy Identify the like terms Group the like terms Simplify 8 I 3 Expanding and Factorising SERIES TOPIC Mathletics Passport 3P Learning

* EXPANDING AND SIMPLIFYING * * How does it work? Your Turn Expanding and Factorising Expanding and simplifying 1 Expand and simplify: a 4^a+ 3h+ a b -3 ^ - xh+ 1.../.../0... c 1p+ 5^p-h d 5d-4^9-3dh e -5b^4- bh+ 3b+ 5b f 9^x-yh- x+ 4y Expand and simplify: a 8( c- 4) + 3( c+ ) b 9^d+ h-( 5-3d) c 3^x-5) - 4 ( + xh d aa ^ + 8) - 5( a + 3h Expanding and Factorising Mathletics Passport 3P Learning I SERIES 3 TOPIC 9

How does it work? Your Turn Expanding and Factorising Expanding and simplifying 3 Expand and simplify: a - ^y+ 4xh-5( x-y) Psst! Remember the 1 can be hidden: - 1^y+ 4xh x^ + x- yh+ x-xy b 3 c a^3+ 4bh+ 4^ab + ah d - 3b^+ bh-^6-bh e d -^- h- ^d-h xy^40x+ 5h - y^10x - xh f 3 -mn^5m- n h+ mn 3 + mn g 3 h ^ h q 4p+ 3q - + q^q+ 5ph 10 I 3 Expanding and Factorising SERIES TOPIC Mathletics Passport 3P Learning

Equations & Inequalities Basics What is a Linear Equation? An equation is a mathematical expression that has two sides separated by an equals sign (=) and at least one variable (or pronumeral ). If the highest power of the variable is 1 then the equation is called a linear equation. For example, these are all equations because = appears in each of them. a x + 4 = 1 b 3x = 15 c + 3x = 8 d x 3 = 5 e 6x 7 = 1 These are also all linear because the power of x (the variable) is 1 (there is no x or x 3 etc). How are Equations Solved? To solve ANY equation the goal is always to get the variable by itself. But whatever is done to find the variable by itself, must also be done to the other side. Each linear equation has only one solution! Finding x by itself If you look at the linear equation x + 4 = 1, it s probably easy to see x = 8. Your brain has actually simplified the equation to have x by itself without you even realising it. This is how: Subtracting 4 from the left side of the equation leaves x by itself. But the same must be done on the right side. x + 4 = 1 x + 4 4 = 1 4 x = 8 For the equation in b, to find x by itself, both sides must be divided by 3 We must do the same to both sides 3x = 15 3x = 15 3 3 x = 5 For the equation in c, it takes two steps to get x by itself Subtract from both sides to leave the term with x by itself Divide both sides by 3 to leave the x by itself + 3x = 8 + 3x = 8 3x = 6 3x = 6 3 3 x = J 6 100% Equations & Inequalities SERIES TOPIC Mathletics 100% 3P Learning

Equations & Inequalities Basics If the variable is in the numerator of a fraction, like in d, then both sides must be multiplied by the denominator. For the equation in d, we need to multiply the left hand side by 3 to find x by itself Sometimes the variable will be in a fraction. Multiply and divide both sides to get the variable by itself. In the equation in e x is the numerator of a fraction x 3 # 3 = 5# 3 x = 15 Multiply both sides by the denominator (always remove the denominator first!) Divide both sides by 6 to leave the x by itself 6x # 7 = 1# 7 7 6x = 84 6x = 84 6 6 x = 14 Sometimes the left side will have to be simplified by collecting like terms: Solve for x in the following linear equation Simplify by collecting like terms Divide both sides by 5 to leave x by itself x+ 5+ 3x = 30 5x + 5 = 30 5x + 5 5 = 30 5 5x = 5 5x = 5 5 5 x = 5 What happens if the Power of the Variable is? If the highest power of the variable is (ie. x is in the equation), then the equation is not linear, but is called a Quadratic equation. To get the variable by itself on one side, the square root is used to change x to x. However, the square root of both sides must be found. Each quadratic equation has two solutions! Solve for x in the following quadratic equation Find the square root of both sides x x + 5 = 1 + 5 5 = 1 5 x x = 16 = ^! 4h x =! 4 So x = -4 or x = 4 Subtract 5 from both sides Because 4 = 16 and ^ 4h = 16 100% Equations & Inequalities Mathletics 100% 3P Learning J 6 SERIES TOPIC 3

Equations & Inequalities Questions Basics 3. Find the value of the variable in each of these linear equations: a x = 8 b 5x = 35 c x 4 = 5 d x 3 = 8 e 3 x = 9 f 5 x = 5 3 100% Equations & Inequalities Mathletics 100% 3P Learning J 6 SERIES TOPIC 5

Equations & Inequalities Questions Basics 4. Solve for the variable in each linear equation: a 4x + 3 = 3 b a - 5 = 9 c 5m + 6 = 31 d -4 + 7n = 4 e -k = 10 f -m + 4 = 6 (Hint: -m = -1m) J 6 6 100% Equations & Inequalities SERIES TOPIC Mathletics 100% 3P Learning

Equations & Inequalities Questions Basics 5. Solve for the variable in each linear equation: a y 4 = 4 b 5x 4 = 15 c p 9 = 6 d 4d 11 = 8 e 7x 3 = 14 f 10m 4 = 5 6. Solve these quadratic equations: a y - 9 = 0 b 4x = 100 100% Equations & Inequalities Mathletics 100% 3P Learning J 6 SERIES TOPIC 7

Equations & Inequalities Knowing More What happens if the Variable is on BOTH sides of the Equation? If the variables appear on both sides of the equation, one side needs to be changed to have no variables in it. REMEMBER! Whatever is done to one side must be done to the other side too. Solve this linear equation: 3x can be subtracted from both sides. Now the right hand side will have no variables. 8x 4 = 3x+ 6 8x 4 3x = 3x+ 6 3x 5x 4 = 6 5x 4+ 4 = 6+ 4 Add 4 to both sides so that 5x is by itself Divide both sides by 5 so that x is by itself 5x = 10 5x = 10 5 5 x = Solve this linear equation: 4p can be added to both sides. Now the right hand side will have no variables. 3p 1= 16 4p 3p 1+ 4p = 16 4p+ 4p 7p 1 = 16 7p 1+ 1 = 16+ 1 Add 1 to both sides so that 7p is by itself Divide both sides by 7 so that p is by itself 7p = 8 7p = 8 7 7 p = 4 If the equation has brackets, they must be expanded first. Then just solve the equation the same way as before. Solve the following linear equation: Expand both brackets 3^x 3h= ^x h 3x 9 = x 4 3x 9 x 4 x x 9 = 4 x can be subtracted from both sides so that the right side will have no variables Add 9 to both sides so that x is by itself x 9+ 9 = 4+ 9 x = 5 J 6 8 100% Equations & Inequalities SERIES TOPIC Mathletics 100% 3P Learning

Equations & Inequalities Questions Knowing More 1. In these linear equations the variable appears on boths sides. Solve for the missing value: a u- 10 = 3u b 7x- 18 = 3x+ 10 c 3^x + h=-3 d 4y+ 18 = 1- y e 10^n- 6h= ^10 + nh f 6m- 4 =- 5^m+ 3h g 8^k- 4h- 5k+ 3 = 4 h 5^y- 1h- 6^y- h+ 3 = 6 i 8t-^t- 18h=-1 j ^a+ 3h- 3^a+ 4h=-10. Find Ivan's mistake when he tried to solve this equation? 3^h+ h= ^h+ 1h+ 5 3h+ = h+ + 5 3h+ = h+ 7 3h+ h = h+ 7 h h = 5 100% Equations & Inequalities Mathletics 100% 3P Learning J 6 SERIES TOPIC 11

Work to be completed at school in September You will complete the following set of work at school in September. Please do not complete it at home.

Date: SIMPLIFYING EXPRESSIONS - RED Simplify the following expressions 1) 5a + 3a ) 6a 4a 3) 4a + a 4) b b 5) 5c 3c + 4c - 7c 6)7c 9c 3c + 8c 7) d 7d d + 10d 8) 6d 4d + 9d - d 9) 7z + z 19z + 10z 10) 8g 7g +g g Student Assessment

Date: SIMPLIFYING EXPRESSIONS - AMBER Simplify the following expressions 1) 4a + 5b + a + 4b ) 5c + 7d + c + d 3) a + 6b a + 4b 4) 5a + 7b + 4a b 5) 9d + 5e 3d e 6) 9f + 10g f 3g 7) 7d 3e + d + 5e 8) 10j 7k j 3k 9) 9h 10g - 7h + g 10) 8m 9n + m - n 11) -5m 6n m 5n 1) 6r -8s 7s + 9r + t 13) 8t 9v + v 9y +3t 14) 4a + 5b 7c 6a b + 4c 15) 3a 6b + c + 4b 9c Student Assessment

Date: SIMPLIFYING EXPRESSIONS - GREEN Simplify the following expressions 1) ab + ba ) 3ab + 4ab ab 3) ab + ba ab + 1 4) 4a 3a 5) 3a + b 3 a + 4b 3 6) a + 3b -5a b + 8 7) a + 5a 3 + 4a + a 3 8) a b + ab + a b 9) ab 4ac + 5ab + 4ac 10) 3cd 4cd + 5cd 11) 7a 3 a 1) 6ab 10ab 3 + 4a b + ab 13) 5d e + e d 10d e + 4e d 14) 4ab -6ab + 7a b 8ab 15) 8a + 4a 9a + 5a 3 Student Assessment

Date: EXPANDING BRACKETS - RED Expand the following 1) 4(a + 3) ) 5(a 6) 3) (3a 4) 4) 5(3a ) 5) 6(a + 5) 6) 10(3a 7) 7) (3a + 5b) 8) 3(a - 7b) 9) 9(a b + 3c) 10) 3(4a 5b + c) Student Assessment

Date: EXPANDING BRACKETS - AMBER Expand the following 1) a(a + 5) ) 3a(4a +7) 3) 5b(3b 8) 4) 4a(3a + 5b) 5) 8a(3a 7b) 6) 10b(6a 5b) 7) ab(a+ 3b) 8) 7a (a + 5b) 9) 4a (a - b) 10) 5ab (a + b) 11) cd(4cd + 3) 1) f g(5f - 6g ) 13) 8ab (a 3 + 6b ) 14) 8a b (ab 1) 15) 9c d 3 (d + e) Student Assessment

Date: EXPANDING BRACKETS - GREEN Expand and simplify 1) 4(3a + b) + (3a + b) ) 5(a + 3b) + 3(a + b) 3) 10(5a + 4b) + 7(5a b) 4) 5(4a 8b) + 6(3a b) 5) 7(a + 5b) 3(a + 3b) 6) 10(a 3b) 4(a + 5b) 7) 5(3a + 4b) 3(a 3b) 8) 10(3a 5b) 6(a b) 9) 3a(a - b) + 5(4a 3b) 10) a(5a + 7b) + 6(a 5b) Student Assessment

Date: SOLVING EQUATIONS - RED Solve the following equations: (1) x + 4 = 10 () x + 7 = 18 (3) x - 5 = 4 (4) x - 7 = 9 (5) x = 14 (6) 4x = 4 (7) x 5 = 9 (8) x 3 = 7 (9) 7x = 56 (10) x 11 = 5 (11) x 5 = 4 (1) 4x = -0 (13) -x + 6 = (14) -5x = 35 Student Assessment

Date: SOLVING EQUATIONS - AMBER Solve the following equations: (1) x 1 = 9 () 3x 4 = 5 (3) 5x + 4 = 4 (4) -10x + 3 = 73 (5) 14 = 8x (6) 87 = 7x + 10 (7) x - 1 = 3 (8) x 4 + = 7 (9) 16 = x 5 + 7 (10) -6x 1 = 11 (11) 70 = -x + 4 (1) - x 5 + 6 = 10 (13) (x + 3) = 1 (14) 4(x + 5) = 4 (15) 5(x - ) = 5 (16) 10(x - 3) = 70 (17) 7(x - 5) = 8 (18) (3x - 1) = 4 Student Assessment

SOLVING EQUATIONS - GREEN (1) 4a + 3 = a + 11 () 6a - = 3a + 10 Date: (3) a 4 = 5a - 19 (4) 9a + 7 = 15a + 1 (5) 5a = a + 4 (6) 3x 6 = 10 - x (7) 9a 3 = 19 a (8) 7a = 17-1a (9) -3a 5 = a + 15 (10) 7 3a = 9 4a (11) 8(a + 1) = (a + 16) (1) 3(5a ) = 4(3a + 6) (13) 4(a + 1) = (8 a) (14) 8(a + 6) = -(3a + 4) Student Assessment

Developing Mathematical Reasoning Task 1 Student Assessment

Developing Mathematical Reasoning Task Student Assessment