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KS3 Mathematics

KS3 Mathematics. A2 Equations. A2 Equations. Contents. A2.2 Equations with the unknown on both sides. A2.1 Solving simple equations. A2.3 Solving more difficult equations. A2.4 Equations and proportion. A2.5 Non-linear equations. Equations.

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KS3 Mathematics

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  1. KS3 Mathematics A2 Equations

  2. A2 Equations Contents A2.2 Equations with the unknown on both sides A2.1 Solving simple equations A2.3 Solving more difficult equations A2.4 Equations and proportion A2.5 Non-linear equations

  3. Equations An equation links an algebraic expression and a number, or two algebraic expressions with an equals sign. For example, x + 7 = 13 is an equation. In an equation the unknown usually has a particular value. Finding the value of the unknown is called solving the equation. x + 7 = 13 x = 6 When we solve an equation we always line up the equals signs.

  4. Using inverse operations 10 – 7 = 3 and 10 – 3 = 7 10 – b = a and 10 – a = b or a = 10 – b and b = 10 – a In algebra, letter symbols represent numbers. Rules that apply to numbers in arithmetic apply to letter symbols in algebra. For example, in arithmetic, if 3 + 7 = 10, we can use inverse operations to write: In algebra, if a + b = 10, we can use inverse operations to write:

  5. Using inverse operations 12 ÷ 4 = 3 and 12 ÷ 3 = 4 12 12 12 12 = a = b a b b a b = a = In arithmetic, if 3 × 4 = 12, we can use inverse operations to write: In algebra, if ab = 12, we can use inverse operations to write: and or and

  6. Using inverse operations to solve equations We can use inverse operations to solve simple equations. For example, x + 5 = 13 x = 13 – 5 x = 8 Always check the solution to an equation by substituting the solution back into the original equation. If we substitute x = 8 back into x + 5 = 13 we have 8 + 5 = 13

  7. Using inverse operations to solve equations Solve the following equations using inverse operations. 5x = 45 17 – x = 6 x = 45 ÷ 5 17 = 6 + x 17 – 6 = x x = 9 11 = x We always write the letter before the equals sign. Check: x = 11 5 × 9 = 45 Check: 17 – 11 = 6

  8. Using inverse operations to solve equations x = 3 7 21 = 3 7 Solve the following equations using inverse operations. 3x – 4 = 14 x = 3 × 7 3x = 14 + 4 3x = 18 x = 21 x = 18 ÷ 3 Check: x = 6 Check: 3 × 6 – 4 = 14

  9. What number am I thinking of…?

  10. What number am I thinking of …? Multiply by 2 and add 5 to give you 11. Start with x x 2x 2x + 5 2x + 5 = 11 I’m thinking of a number. When I multiply the number by 2 and add 5 the answer is 11. What number am I thinking of …? We can write this as an equation. Instead of using ? for the number I am thinking of, let’s use the letter x.

  11. What number am I thinking of …? Multiply by 2 and add 5 to give you 11. Start with x x 2x 2x + 5 2x + 5 = 11 We can solve this equation using inverse operations in the opposite order. Start with the equation: 2x + 5 = 11 Subtract 5: 2x = 11 – 5 2x = 6 Divide by 2: x = 6 ÷ 2 x = 3

  12. Solving equations by transforming both sides m m 4 4 – 1 = 2 ×4 +1 ×4 +1 = 3 Solve this equation by transforming both sides in the same way: Add 1 to both sides. Multiply both sides by 4. m = 12 We can check the solution by substituting it back into the original equation: 12 ÷ 4 – 1 = 2

  13. Find the missing number 23 ? 62 ? ? 127 The number in each brick is found by adding the two numbers above it. What are the missing values? n 23 + n n + 62 We can start by calling the unknown number in the top brick n. The unknown numbers in the next two bricks can be written in terms of n. We can now write an equation and solve it to find the value of n.

  14. Find the missing number The number in each brick is found by adding the two numbers above it. What are the missing values? 23 n 62 21 23 + n n + 62 44 83 127 127 23 + n + n + 62 = 127 Simplify: 85 + 2n = 127 Subtract 85: 2n = 42 Divide by 2: n = 21 Check this solution.

  15. Equation dominoes

  16. A2 Equations Contents A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.3 Solving more difficult equations A2.4 Equations and proportion A2.5 Non-linear equations

  17. Balancing equations

  18. Constructing an equation Ben and Lucy have the same number of sweets. Ben started with 3 packets of sweets and ate 11 sweets. Lucy started with 2 packets of sweets and ate 3 sweets. How many sweets are there in a packet? Let’s call the number of sweets in a packet, n. We can solve this problem by writing the equation: = 3n– 11 2n– 3 The number of Ben’s sweets is the same as the number of Lucy’s sweets.

  19. Solving the equation -2n -2n +11 +11 Let’s solve this equation by transforming both sides of the equation in the same way. Start by writing the equation down. 3n– 11 = 2n– 3 Subtract 2n from both sides. Always line up the equals signs. n– 11 = –3 Add 11 to both sides. This is the solution. n = 8 We can check the solution by substituting it back into the original equation: 3  8 – 11 = 2  8 – 3

  20. Constructing an equation = 4n n + 9 I’m thinking of a number. When I multiply the number by 4, I get the same answer as adding 9 to the number. What number am I thinking of? Let’s call the unknown number n. We can solve this problem by writing the equation: The number multiplied by 4 is the same as the number plus 9.

  21. Solving the equation -n -n ÷3 ÷3 Let’s solve this equation by transforming both sides of the equation in the same way. 4n = n + 9 Start by writing the equation down. Subtract n from both sides. 3n = 9 Always line up the equals signs. Divide both sides by 3. n = 3 This is the solution. We can check the solution by substituting it back into the original equation: 4  3 = 3 + 9

  22. Constructing an equation (2x + 5)o (65 – 2x)o Find the value of x. Remember, vertically opposite angles are equal. We can solve this problem by writing the equation: 2x + 5 = 65 – 2x

  23. Solving the equation +2x +2x -5 -5 ÷4 ÷4 Let’s solve this equation by transforming both sides of the equation in the same way. 2x + 5 = 65 – 2x Add 2x to both sides. 4x + 5 = 65 Subtract 5 from both sides. 4x = 60 Divide both sides by 4. x = 15 This is the solution. Check: 2  15 + 5 = 65 – 2  15

  24. Rectangle problem The area of this rectangle is 27 cm2. Calculate the value of x and use it to find the height of the rectangle. Opposite sides of a rectangle are equal. 8x – 14 2x + 1 We can use this fact to write an equation in terms of x.

  25. Rectangle problem ÷6 ÷6 +14 –2x +14 –2x The area of this rectangle is 27 cm2. 8x – 14 = 2x + 1 6x – 14 = 1 8x – 14 2x + 1 6x = 15 x = 2.5 If x = 2.5 we can find the height of the rectangle using substitution: 8 × 2.5 – 14 = 20 – 14 = 6 cm

  26. Rectangle problem ÷6 ÷6 The area of this rectangle is 27 cm2. What is its width? y Let’s call the width of the rectangle y. If the height of the rectangle is 6 cm and the area is 27 cm2 then we can find the width by writing the equation: 8x – 14 2x + 1 6y = 27 The dimensions of the rectangle are 6 cm by 4.5 cm. y = 4.5

  27. Select the correct equation Thomas has £1 and buys 7 chocolate bars. Veronica has 58p and buys 4 chocolate bars. They both receive the same amount of change. If c is the cost of one chocolate bar, which equation could we use to solve this problem? A: 4c + 58 = 7c + 100 B: 58 – 4c = 1 – 7c C: 58 – 4c = 100 – 7c C: 58 – 4c = 100 – 7c D: 4c– 58 = 7c – 1

  28. Equation match

  29. A2 Equations Contents A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.3 Solving more difficult equations A2.4 Equations and proportion A2.5 Non-linear equations

  30. Equations with brackets ÷ 2 - 4x + 10 ÷ 2 - 4x + 10 Equations can contain brackets. For example: 2(3x– 5) = 4x To solve this we can Multiply out the brackets: –10 6x = 4x Add 10 to both sides: 6x = 4x + 10 Subtract 4x from both sides: 2x = 10 Divide both sides by 2: x = 5

  31. Equations with brackets + 5 - 2x + 5 - 2x Sometimes we can solve equations such as: 2(3x– 5) = 4x by first dividing both sides by the number in front of the bracket: Divide both sides by 2: 3x – 5 = 2x Add 5 to both sides: 3x = 2x + 5 Subtract 2x from both sides: x = 5 In this example, dividing first means that there are fewer steps.

  32. Balancing equations with brackets

  33. Solving equations involving division 3x + 2 = 11 –x 4 Linear equations with unknowns on both sides can also involve division. For example, In this case we must start by multiplying both sides of the equation by 4. 3x + 2 = 4(11 –x) Multiply out the brackets: 3x + 2 = 44 – 4x Add 4x to both sides: 7x + 2 = 44 Subtract 2 from both sides: 7x = 42 Divide both sides by 7: x = 6

  34. Solving equations involving division 4 5 = (x + 3) (3x – 5) Sometimes the expressions on both sides of the equation are divided. For example, In this example, we can multiply both sides by (x + 3) and (3x– 5) in one step to give: 4(3x– 5) = 5(x + 3) Multiply out the brackets: 12x– 20 = 5x + 15 Subtract 5x from both sides: 7x – 20 = 15 Add 20 to both sides: 7x = 35 Divide both sides by 7: x = 5

  35. Equivalent equations

  36. A2 Equations Contents A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.4 Equations and proportion A2.3 Solving more difficult equations A2.5 Non-linear equations

  37. Sale! Original Price £4 Sale Price £3 sale price We can write the ratio of for each pair of values. original price 12 15 6 3 9 16 20 8 4 12 Footballs! Were £4, Now only £3! £8 £12 £16 £20 £6 £9 £12 £15 = = = = = 0.75 The ratios are equal. What do you notice about these ratios?

  38. Drawing a graph Original Price, x £4 £8 £12 £16 £20 Sale Price, y £3 £6 £9 £12 £15 3 y = x 4 y Sale price of footballs The points lie on a straight line through the origin. 15 12 The equation of the line is: 9 Sale price, £ 6 or 3 y = 0.75x 0 x 4 8 12 16 20 0 Original price, £

  39. Constant speed Time, minutes 10 Distance, metres 800 distance We can write the ratio of for each pair of values. time 160 320 480 640 800 2 4 6 8 10 Susan walks to school at a constant speed. Altogether, it takes her 10 minutes to walk 800 metres. How far would she have walked in two minutes? 2 4 6 8 160 320 480 640 = = = = = 80 The ratios are equal. What do you notice about these ratios? The time and the distance are directly proportional.

  40. Distance/time graph Time, minutes 2 4 6 8 10 Distance, metres 160 320 480 640 800 y 800 640 480 Distance, metres 320 160 0 x 2 4 6 8 10 0 Time, minutes We can plot the points from the table onto a graph. The points lie on a straight line through the origin. Susan’s walking speed The equation of the line is: y = 80x How far would Susan walk in: a) 3 minutes? 240 metres a) 15 minutes? 1200 metres

  41. Mixing colours 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint Orange paint is made by mixing 4 litres of yellow paint with 6 litres of red paint. To make the same shade of orange we must keep the amount of yellow paint and red paint in direct proportion. How many litres of each colour do you need to make: 2 l of yellow and 3 l of red a) 5 litres of orange paint? b) 1 litre of orange paint? 0.4 l of yellow and 0.6 l of red c) 7 litres of orange paint? 2.8 l of yellow and 4.2 l of red

  42. Mixing colours red 6 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint The ratio is = 1.5 yellow 4 We can write an equation linking the amount of red paint r to the amount of yellow paint y as r = 1.5y How many litres of red paint would be needed to mix with 14 litres of yellow paint to make the same shade of orange? r = 1.5 × 14 = 21 l

  43. Mixing colours

  44. Equations and direct proportion y where k is a constant value equal to . x When two quantities x and y are directly proportional to each other we can link them with the symbol . y is directly proportional to x. We write y x We can also link these variables with the equation y = kx The graph of y = kx will always be a straight line through the origin.

  45. A2 Equations Contents A2.1 Solving simple equations A2.2 Equations with the unknown on both sides A2.5 Non-linear equations A2.3 Solving more difficult equations A2.4 Equations and proportion

  46. Non-linear equations A number added to its square equals 42. What could the number be? If we call the unknown number x we can write the following equation: x + x2 = 42 This is a non-linear equation. It contains powers greater than 1. One solution to this equation is x = 6. There is another solution to this equation. x = –7 6 + 62 = 6 + 36 = 42 and –7 + (–7)2 = –7 + 49 = 42

  47. Non-linear equations In a linear equation, unknowns in the equation cannot be raised to any power other than 1. For example, 4x + 7 = 23 is a linear equation. In a non-linear equation, unknowns in the equation can have indices other than 1. For example, x4 + 2x = 20 is a non-linear equation. In a quadratic equation, the highest index of any of the unknowns is 2. For example, x2 – 3x = 10 is a quadratic equation. Quadratic equations usually have two solutions.

  48. Solving non-linear equations Solve 3x2 – 5 = 22 We can solve this non-linear equation using inverse operations. 3x2 – 5 = 22 Adding 5: 3x2 = 27 Dividing by 3: x2 = 9 Square rooting: x = ±9 x = 3 or –3 When we find the square root in an equation we must always give both the positive and the negative solution.

  49. Solving non-linear equations 25 Solve m – 6 = m– 6 We must start by multiplying by m – 6. (m – 6)(m – 6) = 25 We can write this as: (m – 6)2 = 25 Square rooting: m – 6 = 5 or –5 To find both solutions we must add 6 to both 5 and –5. or m = 5 + 6 m = –5 + 6 m = 11 m = 1

  50. Using a calculator to solve non-linear equations 25 m– 6  3 n = 7 3 We know that 7 is not a cube number so we can use the key on a calculator. Solve 4n3 + 5 = 33 We can solve this equation by using inverse operations. 4n3 + 5 = 33 Subtracting 5: 4n3 = 28 Dividing by 4: n3 = 7 Cube rooting: n = 1.91 (to 2 d.p.)

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