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  • Level: GCSE
  • Subject: Maths
  • Word count: 2709

algebra coursework

Extracts from this document...

Introduction

Algebra Coursework

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Ex 1

2 X 2 square on 10 X 10 grid

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12 X 23 = 276
13 X 22 = 286

286 – 276 = 10

Z

Z+1

Z+10

Z+11

Z = Top left number = 12 (in this case)
Z+1 = 12 +1 = 13 (top right number)
Z+10 = 12+10 = 22 ( bottom left number)
Z+11= 12+11 = 23 ( bottom right  number)

Z (Z+11) = Z² + 11Z

(Z+1) (Z+10) = Z² +10Z + Z +10
                      = Z² + 11Z + 10

Z² + 11Z + 10 – Z² + 11Z = 10

Ex 2

2 X 2 square on 10 X 10 grid

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74 X 85 = 6290
84 X 75 = 6300

6300 – 6290 = 10

Z

Z+1

Z+10

Z+11

Z = Top left number = 74 (in this case)
Z+1 = 74+1 = 75 ( top right number )
Z+10 = 74+10 = 84 ( bottom left number)
Z+11= 74+11 = 85 ( bottom right number)

Z (Z+11) = Z² + 11Z

(Z+1) (Z+10) = Z² +10Z + Z +10
                      = Z² + 11Z + 10

Z² + 11Z + 10 – Z² + 11Z = 10

Ex 3

3 X 3 square on 10 X 10 grid

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48 X 70 = 3360
68 X 50 = 3400

3400 – 3360 = 40

Z

Z+2

Z+20

Z+22

Z = top left number = 48 (in this case)

Z+2 = 48 +2 = 50 (top right number)

Z+20 = 48 +20 = 68 (bottom left number)

Z+22= 48 +22 = 70 (bottom right number)

Z (Z+22) = Z² + 22Z

(Z+2) (Z+20) = Z² + 20Z +2Z +40

                      = Z² + 22Z + 40

Z² + 22Z + 40 - Z² + 22Z = 40

Ex 4

4 X 4 square on 10 X 10 grid

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7 X 40 = 280

10 X 37 = 370

370 – 280 = 90

Z

Z+3

Z+30

Z+33

Z = top left number = 7 (in this case)

Z+3 = 7 + 3 = 10 (top right number)

Z+30 = 7 + 30 = 37 (bottom left number)

Z+33 = 7+33 = 40 (bottom right number)

Z (Z+33) = Z² + 33 Z

(Z+3) (Z+30) = Z² + 30Z + 3Z + 90

                      = Z² + 33Z + 90

Z² + 33Z + 90 - Z² + 33Z = 90

Ex 5

5 X 5 square on 10 X 10 grid

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32 X 76 = 2432

36 X 72 = 2592

2592 – 2432 = 160

Z

Z+4

Z+40

Z+44

Z = top left number = 32 (in this case)

Z+4 = 32 + 4 = 36 (top right number)

Z+40 = 32 + 40 = 72 (bottom left number)

...read more.

Middle

Z (Z+23) = Z² +23Z

(Z+3) (Z+20) = Z² + 20Z +3Z +60

                      = Z² + 23Z +60

Z² + 23Z +60 - Z² +23Z = 60

Ex 10

3 X 5 rectangle on a 10 X 10 grid

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41 X 65 = 2665

45 X 61 = 2745

2745 – 2665 = 80

Z

Z+4

Z+20

Z+24

Z (Z+24) = Z² +24Z

(Z+4) (Z+20) = Z² +20Z +4Z +80

                       = Z² + 24Z +80

Z² + 24Z +80 - Z² +24Z = 80

Rectangle formula for a 10 X 10 grid

                                                     Y = 3

Z

Z+(Y-1)

Z-10+10X

Z-10+10X+Y-1

                                      X = 2

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Z = 1 (in this case, it is the top left number)

Z+(Y-1) = 3 (top right number)

Z+ (3-1) =3

Z-10 +10X = 11 (bottom left number)

Z-10 +10 x 2 = 11

Z-10+10X+Y-1 = 13 (bottom right number)

 Z-10 +10 x 2 + 3 – 1 = 13

Z (Z-10+10X+Y-1) = Z (Z-11+10X+Y)

                                = Z² - 11Z + 10XZ +YZ

(Z-10+10X) (Z+Y-1) = Z² + YZ –Z -10Z – 10Y +10 +10XZ +10XY -10X

Z² + YZ –Z -10Z – 10Y +10 +10XZ +10XY -10X - Z² - 11Z + 10XZ +YZ

= -10Y +10 +10XY -10X

Formula = 10(Y-1) (X-1)

-10Y +10 +10XY – 10X ÷ 10 = -Y +1 +XY –X

-Y +1 +XY –X

(X-1) (Y-1)

10 (X-1) (Y-1) = -10Y +10 +10XY -10X

I worked out the formula 10(X-1) (Y-1) by multiplying  

Z (Z-11+10X+Y) + (Z-10+10X) (Z+Y-1)

Then took them away from eachother.

Z² + YZ –Z -10Z – 10Y +10 +10XZ +10XY -10X - Z² - 11Z + 10XZ +YZ

To get -10Y +10 +10XY – 10X

Which factorises to give 10(Y-1) ( X-1)

Ex 11

3 X 5 rectangle on 10 X 10 grid = 80

10(Y-1) (X-1)  

10(5-1) (3-1) = 80

2 X 4 rectangle on 10 X 10 grid = 30

10(Y-1) (X-1)

10 (4-1) (2-1) = 30

  The formula 10(Y-1) (X-1) works for rectangles on a 10 X 10 grid          

Rectangle formula for a rectangle on any size number grid

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The formula for any size rectangle on a 10X10 grid is 10(Y-1) (X-1) therefore to find out the formula for a rectangle on any size number grid we simply have to increase or decrease the number but this would not be a formula.

9X9 number grid

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The formula for this grid would be 9(Y-1) (X-1)

EX 12

Y =3

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X =2

1 X 12 = 12
3 X 10 = 30

30 – 12 = 18

9(Y-1) (X-1)

9(3-1) (2-1) =18

General formula for a rectangle on any size grid explanation

                                                     W

Y

…………..

………….

…………..

………….

X

................

…………..

                                      W

W represents the size of number grid

X and Y represent the sides of the rectangle

The general formula for any rectangle on any number grid = W(Y-1) X-1)

Ex 13

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1 X 12 = 12
3 X 10 = 30

30 – 12 = 18

Z

Z+2

Z+9

Z+11

Z = top left number (1 in this case)

Z+2 = 3 (top right number) 1+2 = 3

Z+9 = 10 (bottom left number) 1+9 = 10

Z+11= 12 (bottom right number) 1+11 = 12

Z (Z+11) = Z² + 11Z

(Z+9) (Z+2) = Z² + 2Z + 9Z + 18

                    = Z² + 11Z + 18

Z² + 11Z + 18 - Z² + 11Z = 18

9(Y-1) (X-1)

9(3-1) (2-1) =18

Ex 14

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...read more.

Conclusion

Z = 1 (in this case)

Z+Y-1 = 3 (1+ 3 -1 = 3)

Z-W+10X = 11 (1-10+20 = 11)

Z-W+10X+Y-1 = 13 (1-10+20+3-1 = 13)

Z (Z-W+10X+Y-1) = Z² - ZW + 10XZ + ZY – Z

(Z-W+10X) (Z+Y-1) = Z² + ZY – Z – ZW – YW – W + 10XZ + 10 XY – 10 X

 (Z² + ZY – Z – ZW – YW – W + 10XZ + 10 XY – 10 X) – (Z² - ZW + 10XZ + ZY – Z)

= -WY – W + 10XY – 10X

-WY – W + 10XY – 10X ÷ 10 = -WY – W + XY – X

Which factorises to give W(Y-1) (X-1)

The formula at work

4 X 3 rectangle on a 5 X 5 grid = 30

5X5

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4 X 3 rectangle

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1 X 14 = 14

4 X 11 = 44

44-14 = 30

Z

Z+3

Z+10

Z+13

Z=1 (in this case) top left number

Z+3 = 4 (top right number) 1+3 = 4

Z+10 = 11 (bottom left number) 1+10 = 11

Z+13 = 14 (bottom right number) 1+13= 14

Z (Z+13) = Z² + 13Z

(Z+3) (Z+10) = Z² + 10Z + 3Z+ 30

                      = Z² + 13Z + 30

Z² +13Z + 30 - Z² + 13Z = 30

W(Y-1) (X-1)

5(4-1) (3-1) = 30

7X7

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4X2

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4X14 = 56

7X11 = 77

77-56 = 21

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Z+3

Z+7

Z+10

Z = 4 (in this case)

Z+3 = 7 (top right number) 4 + 3 = 7

Z+7 = 11 (bottom left number) 4 + 7 = 11

Z+10 = 14 (bottom right number) 4 + 10 = 14

Z (Z+10) = Z² + 10Z

(Z+3) (Z+7) = Z² + 7Z +3Z +21

                     = Z² + 10Z + 21

Z² + 10Z + 21 - Z² + 10Z = 21

W(Y-1) (X-1)

7(4-1) (2-1) = 21

Further investigation

Further steps could be taken in the investigation process, unfortunately time is not on our side. For example the shape of the grid could have been altered to a rectangle and then a formula for a certain rectangle could be found, then a formula for any rectangle could be found.

...read more.

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