Equations ppt
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1. Let x = Archibald's marbles. Then Muriel has 2x marbles and Oswald has 95 - x - 2x marbles. 2. Set up an equation where their total marbles equals 95: x + 2x + (95 - x - 2x) = 95 3. Solve the equation for x. Archibald has 25 marbles. 4. Check that with x = 25, their individual amounts add to 95.
Equations ppt
- 1. Equations An equation is any mathematical statement that contains an = sign. 6 + 4 = 10 5 – 9 = – 4 8 + 4 = 15 – 3 6 x 4 = 24 27 3 = 9 y + 4 = 10 2a + 4 = 12 are all examples of equations!
- 2. If we begin with a true equation 9 + 6 = 15 We can do anything we like (add, multiply, subtract or divide) to the numbers on either side of the = sign as long as we do the same thing to both sides!
- 3. 9 + 6 = 15 Begin with a true statement Add 3 to both sides 9 + 6 + 3 = 15 + 3 18 = 18 still true! Subtract 8 from both sides 9 + 6 – 8 = 15 – 8 7 = 7 still true! Multiply both sides by 5 (9 + 6) x 5 = 15 x 5 75 = 75 still true! Divide both sides by 3 (9 + 6) 3 = 15 3 5 = 5 still true!
- 4. This is a very useful process when there is an unknown (like x) on one side, and we wish to isolate it to solve the equation. Solve x – 9 = 5 We want x on its own on the left of the = sign, so we aim to get rid of the – 9. The opposite of – 9 is + 9, so we ADD 9 to both sides x – 9 + 9 = 5 + 9 x = 14x = 5 + 9 ZERO!
- 5. Example 1: Solve y – 5 = – 3 We need to remove the 5 by “undoing” the minus. The opposite of minus is add, so we ADD 5 to both sides : We can now cancel the 5s on the left side, and at the same time work out – 3 + 5 y = 2 y – 5 = – 3+ 5 + 5
- 6. Example 2: Solve 5y = 30 As this is really 5 times y = 30, to isolate the y we need to remove the 5 by “undoing” the times. The opposite of times is divide, so we DIVIDE both sides by 5: 5 30 5 5 y We can now cancel the 5s on the left side, and at the same time work out 30 5 y = 6
- 7. Example 3: Solve y/3 = 12 As this is really y divided by 3 = 12, we need to remove the 3 by “undoing” the divide. The opposite of divide is multiply, so we MULTIPLY both sides by 3: 12 3 y We can now cancel the 3s on the left side, and at the same time work out 12 x 3 y = 36 x3x3
- 8. EXAMPLE 4: Solve 3y + 5 = 26 To get x alone, first we need to remove the 5, then the 3. Begin by taking 5 from both sides 3y + 5 = 26 3y = 21 Now we divide both sides by 3 y = 7 ZERO – 5– 5 3 21 3 3 y Cancel the 3s on the left side
- 9. And now for a really useful trick! Suppose we begin with 8 – 3 = 5 You’re allowed to change all the signs (the sign in front of every term) Still true! Try this for 2 – 9 – 5 = – 12 and get – 2 + 9 + 5 = + 12 Still true! – 8 + 3 = – 5 But remember you must change ALL the signs
- 10. This trick is really useful in equations where there is a negative in front of the letter! Solve – a + 7 = 12 Change all the signs Now add 7 to both sides, as before and get a – 7 = – 12 a – 7 + 7 = – 12 + 7 a = – 5 Now check to see you’re right by substituting a = – 5 into the original equation – – 5 + 7 = 12 TRUE!
- 11. Example 6 Solve 5 – 2a = – 9 Change all the signs as the “a” has a minus in front Now add 5 to both sides, as before – 5 + 2a = 9 – 5 + 2a + 5 = 9 + 5 2a = 14 Now check to see you’re right by substituting a = 7 into the original equation 5 – 2 x 7 = – 9 TRUE! a = 7
- 12. Solve 7 4 a Since this is the same as a 4 = 7, we do the opposite of divide, i.e. multiply by 4 a = 28 x 4 x 4 4 a = 7 Cancel the 4s on the left
- 13. Example 8: Solve 6 5 37 a First we multiply by 5 to get rid of the fraction 6 5 37 a x 5 x 5 Cancel the 5s on the left 7 – 3a = 30 Seeing there’s a minus in front of the a, we can change all signs – 7 + 3a = – 30 Add 7 to both sides – 7 + 3a + 7 = – 30 + 7 3a = – 23 Divide both sides by 3 a = – 7.667
- 14. Example 9: Solve 6 5 3 4 a First we subtract 4 from both sides Cancel the 4s on the left Multiply both sides by 5 3a = 10 Divide both sides by 3 a = 10/3 2 5 3a 464 5 3 4 a – 4 – 4
- 15. Example 10: Solve 4 7 2 9 a Add 9 to both sides Multiply both sides by 7 Divide both sides by 2 a = 45.5 13 7 2a Sign change 4 7 2 9 a 2a = 91
- 16. Solve 2(x – 5) = 15 Expand the brackets 2x – 10 = 15 Add 10 to both sides 2x – 10 + 10 = 15 + 10 2x = 25 Divide both sides by 2 x = 25/2 or 12.5 Example 11……….
- 17. Solve 2(x – 5) + 3(2x + 1) = 28 Expand the brackets 2x – 10 + 6x + 3 = 28 Add 7 to both sides 8x – 7 + 7 = 28 + 7 8x = 35 Divide both sides by 8 x = 35/8 or 4.375 Example 12………. Clean up left side 8x – 7 = 28
- 18. Equations with an unknown on both sides Example 13 Solve 3a – 5 = a + 11 The aim is to get the a on one side only, so try taking a from both sides: What happens to the right-hand side? 3a – 5 – a = a + 11 – a 3a – 5 – a = a + 11 – a 11 Add 5 to both sides 2a – 5 + 5 = 11 + 5 2a = 16 Divide both sides by 2 a = 8 2a – 5 = Back at the start, we could have taken 3a from both sides instead of just a. This would have given – 5 = 11 – 2a and then proceed as usual!
- 19. Equations with an unknown on both sides Example 14 Solve 9 – a = 12 + 3a The aim is to get only one term with a. So try adding a to both sides: What happens to the left-hand side? 9 - a + a = 12 + 3a + a 9 – a + a = 12 + 3a + a 12 + 4a Take 12 from both sides 9 – 12 = 12 + 4a – 12 – 3 = 4a Divide both sides by 4 a = – ¾ 9 =
- 20. Equations with fractions on both sides Example 15 Solve 4 2 3 13 xx Multiply both sides by the LCD, 12. This kills the fractions. Put brackets around the numerators. 1 12 4 )2( x 1 12 3 )13( x Cancel 4(3x + 1) = 3(2 – x) Expand 12x + 4 = 6 – 3x 15x = 2 x = 2/15 4 3
- 21. Equations with fractions on both sides Example 16 Solve 4 5 5 )24(3 xx Multiply both sides by the LCD, 20. This kills the fractions 1 20 4 5 x 1 20 5 )24(3 x Cancel and make sure brackets are around numerators 4 3(4 – 2x) =5(5 - x) Expand 48 – 24x = 25 – 5x 23 = 19x x = 23/19
- 23. Example 17 A rectangular field is 5m longer than it is wide Its perimeter is 200m. Find its dimensions (width and length). Key Strategy ….. Always let x equal the smallest part So, Let x equal the width. So the length is… x + 5 and The four sides total to 200………….. x + x + x + 5 + x + 5 = 200 4x + 10 = 200 4x = 190 x = 47.5 So the width is 47.5m Length is 47.5 + 5 = 52.5m Finally make sure they add to 200 x x + 5 Instead of writing x + 5 twice, you could have written 2(x + 5). This becomes 2x + 10 when you get rid of the brackets!
- 24. Example 18 Another rectangular field is 12m longer than it is wide Its perimeter is 1km. Find its dimensions (width and length). Let x equal the width. So the length is… x + 12 and The four sides total to 1000………….. 4x + 24 = 1000 4x = 976 x = 244 So the width is 244m Length is 244 + 12 = 256m Finally make sure they add to 1000 x x + 12
- 25. Example 19 Find the value of x in this diagram (4x – 3)º (7x – 4)º As these are co- interior, they are supplementary and so must add to 180º 4x – 3 + 7x – 4 = 180 Clean up left side 11x – 7 = 180 Add 7 to both sides 11x = 187 Divide both sides by 11 x = 17 It is wise to check your answer by substituting 17 into both angles and seeing that they add to 180. 7 x 17 – 4 = 115 4 x 17 – 3 = 65 115 + 65 = 180, so we’re correct!
- 26. Example 18 Find the value of b in this isosceles triangle (2b + 1)º (b – 7 )º As it’s isosceles, the other bottom angle must also be (2b + 1) The three angles add to 180, so….. b – 7 + 2(2b + 1) = 180 b – 7 + 4b + 2 = 180 5b – 5 = 180 5b = 185 b = 37 Now check your answer by substituting 37 into the 3 angles and seeing that they add to 180. 37 – 7 = 30 2 x 37 + 1 = 75 30 + 2 x 75 = 180, so we’re correct! (2b + 1)º Expanding brackets Cleaning up left side Adding 5 to both sides Dividing both sides by 5
- 27. Example 19 Jimmy, Mary and Joseph have $24 between them. Mary has twice the amount Jimmy has. Joseph has $3.25 more than Mary. How much do they each have? Key Strategy ….. Always let x equal the smallest share So, Let x equal Jimmy’s amount as he has the least. So Mary has……… 2x and Joseph has………. 2x + 3.25 Now we know they total to 24………….. x + 2x + 2x + 3.25 = 24 5x + 3.25 = 24 5x = 20.75 x = 4.15 So Jimmy has $4.15 Mary has 2 x $4.15 = $8.30 Joseph has $8.30 + $3.25 = $11.55Finally make sure they add to $24
- 28. Mary is twice as old as John, and 4 years younger than Peter. The sum of their ages is 159. How old are they? Let the youngest (John) be x. So Mary’s age is 2x & Peter’s age is 2x + 4 Now we add them up, knowing it will equal 159. x + 2x + 2x + 4 = 159 5x + 4 = 159 5x = 155 x = 31 So John is 31 Mary is 2 x 31 = 62 Peter is 62 + 4 = 66– 4 from both sides divide both sides by 5 Example 20
- 29. The isosceles triangle and the square have the same perimeter. Find x as a mixed numeral 2x – 3 x + 5 Triangle’s perimeter = x + 5 + 4x + 2 = x + 5 + 2(2x + 1) = 5x + 7 Square’s perimeter = 4(2x - 3) = 8x - 12 5x + 7 = 8x – 12 5x + 7 – 5x = 8x – 12 – 5x 7 = 3x – 12 7 + 12 = 3x – 12 + 12 19 = 3x 3 1 6x Example 21
- 30. Twins Bessie and Albert have a brother, Marmaduke, 8 years older than they are, and they have a sister, Sylvia, who is 12 years younger than they are. Together their ages add to 168. Use algebra to find the twins’ ages. Let the twins’ ages be x. Marmaduke is x + 8. Sylvia is x – 12. x + x + x + 8 + x – 12 = 168 4x – 4 = 168 4x = 168 + 4 4x = 172 x = 43 The twins are 43! Also, Marmaduke is 51, Sylvia is 31
- 31. (3a – 5)cm Robbie the rectangle is twice as long as he is wide. His perimeter is 294 cm. Calculate his dimensions and his area. 6(3a – 5) = 294 18a – 30 = 294 18a = 324 a = 18 Width = 49cm Length = 98cm Area: 49 x 98 = 4802cm2 If his width is (3a – 5) then his length is twice that, so must be 2(3a – 5). This means all sides must total 6(3a – 5) (3a – 5) 2(3a – 5) 2(3a – 5)
- 32. Archibald, Muriel and Oswald come across a bag of 95 marbles. They divide them up in such a way that Muriel has 50% more than Archibald, and Oswald has five fewer than Archibald and Muriel combined. How many does each have? Let x be Archibald’s share So Muriel’s share is 1.5x Oswald’s share is x + 1.5x – 5 = 2.5x – 5 x + 1.5x + 2.5x – 5 = 95 5x – 5 = 95 5x = 100 x = 20 Archie has 20, Muriel 30 and Ossie 45
- 33. Example 25 5x - 4 3x + 2 20 – x The perimeter of this shape is 176 cm. Find the value of x This side is 5x – 4 – (3x + 2) 2x - 6 And so is this side also 2x - 6 = 5x – 4 – 3x – 2 = 2x - 6 So this side has to be 20 – x + 2x – 6 = x + 14 x + 14 Now add up all the sides! 12x + 20 = 176 12x = 156 X = 13
- 34. Attila, Otto, Peregrine and Ugly are cousins. Peregrine is two decades younger than Ugly, and Peregrine’s age is 80% of Attila’s age. Otto, the eldest, is 26 years younger than the total of Attila’s and Peregrine’s ages. Their ages total eight less than four times Ugly’s age. How old are they? Example 26
- 35. Try letting Attila’s age = x, only because it says “Peregrine’s age is 80% of Attila’s age” making it easy to write Peregrine’s age as 0.8x. So….. Let Attila’s age = x Peregrine’s age = 0.8x Ugly’s age = Peregrine’s age + 20 = 0.8x + 20 Otto’s age = Attila + Peregrine - 26 = x + 0.8x – 26 = 1.8x – 26
- 36. Their ages total eight less than four times Ugly’s age. Attila = x Peregrine = 0.8x Ugly = 0.8x + 20 Otto = 1.8x – 26 x + 0.8x + 0.8x + 20 + 1.8x – 26 = 4(0.8x + 20) - 8 4.4x - 6 = 3.2x + 80 - 8 1.2x = 78 x = 65 Attila is 65 Peregrine is 52 Ugly is 72 Otto is 91
- 37. Little Jimmy has a number of 10c and 20c coins in his piggybank. His 29 coins total to $4.10, How many of each kind of coin does he have? Let x be the number of 10 cent coins. Then, since there are 29 coins altogether, we can let the number of 20c coins be 29 – x . Example 27
- 38. so we now have that there are…. • x coins each valued at 10c, and… • (29 – x) coins each valued at 20c The x coins each valued at 10 cents must be worth a total of 10x and the (29 – x) coins each valued at 20 cents must be worth a total of 20(29 – x) We know these values total to 410, so 10x + 20(29 – x) = 410 10x + 580 – 20x = 410 580 – 10x = 410 Expand Clean up
- 39. – 10x = 410 – 580 – 10x = – 170 x = 17 Remember, x was the number of 10 cent coins, so there are 17 ten-cent coins. There were 29 coins altogether, so there must be (29 – x) i.e.29 – 17 = 12 twenty-cent coins! Clean up Divide by – 10 Number of 10c = 17 Number of 20c = 12 Finally, check that 17 x 10 + 12 x 20 = 410
- 40. Example 28 – This uses FACTORISING The diagram represents a path enclosing a park. The curved section is a quadrant of a circle, radius r m. The longest side is twice the width of the park. The perimeter is 1km. Calculate the area of the park in m2 Circumference of a circle C = 2π r Area of a circle A = π r 2
- 41. r rr r r ¼ x 2πr We set up an equation for the perimeter…. Left side + top + bottom + quadrant = 1000m r + r + 2r + = 1000mr 2 4r + = 1000mr 2 1000 2 4 r 2 4 1000 r r = 179.50755m Area = ¼ π r 2 + r 2 Area = 57531 sq metres Factorising!