Equations An equation is any mathematical statement that contains an = sign. 6 x 4 = 24 6 + 4 = 10 8 + 4 = 15 – 3 5 – 9 = – 4 27  3 = 9 y + 4 = 10 are all examples of equations! 2a + 4 = 12

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!

Begin with a true statement 9 + 6 = 15 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!

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 ZERO! x – 9 + 9 = 5 + 9 x = 14 x = 5 + 9

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 : y – 5 = – 3 + 5 + 5 We can now cancel the 5s on the left side, and at the same time work out – 3 + 5 y = 2

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: We can now cancel the 5s on the left side, and at the same time work out 30 ÷ 5 y = 6

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: x3 x3 We can now cancel the 3s on the left side, and at the same time work out 12 x 3 y = 36

y = 7 EXAMPLE 4: Solve 3y + 5 = 26 3y + 5 = 26 3y = 21 To get x alone, first we need to remove the 5, then the 3. Begin by taking 5 from both sides ZERO 3y + 5 = 26 – 5 – 5 3y = 21 Now we divide both sides by 3 Cancel the 3s on the left side y = 7

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) – 8 + 3 = – 5 Still true! Try this for 2 – 9 – 5 = – 12 and get – 2 + 9 + 5 = + 12 Still true! But remember you must change ALL the signs

This trick is really useful in equations where there is a negative in front of the letter! Example 5 Solve – a + 7 = 12 Change all the signs a – 7 = – 12 Now add 7 to both sides, as before a – 7 + 7 = – 12 + 7 a = – 5 and get Now check to see you’re right by substituting a = – 5 into the original equation – – 5 + 7 = 12 TRUE!

Example 6 Solve 5 – 2a = – 9 Change all the signs as the “a” has a minus in front – 5 + 2a = 9 Now add 5 to both sides, as before – 5 + 2a + 5 = 9 + 5 2a = 14 a = 7 Now check to see you’re right by substituting a = 7 into the original equation 5 – 2 x 7 = – 9 TRUE!

a = 28 Example 7 Solve Since this is the same as a  4 = 7, we do the opposite of divide, i.e. multiply by 4 = 7 x 4 x 4 Cancel the 4s on the left a = 28

Example 8: Solve First we multiply by 5 to get rid of the fraction 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

Example 9: Solve First we subtract 4 from both sides – 4 Cancel the 4s on the left Multiply both sides by 5 3a = 10 Divide both sides by 3 a = 10/3

Example 10: Solve Sign change Add 9 to both sides Multiply both sides by 7 2a = 91 Divide both sides by 2 a = 45.5

Brackets x = 25/2 or 12.5 Example 11………. 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 12………. Brackets x = 35/8 or 4.375 Solve 2(x – 5) + 3(2x + 1) = 28 Expand the brackets Brackets 2x – 10 + 6x + 3 = 28 Clean up left side 8x – 7 = 28 Add 7 to both sides 8x – 7 + 7 = 28 + 7 8x = 35 Divide both sides by 8 x = 35/8 or 4.375

and then proceed as usual! 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: 3a – 5 – a = a + 11 – a What happens to the right-hand side? 3a – 5 – a = a + 11 – a 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! 2a – 5 = 11 Add 5 to both sides 2a – 5 + 5 = 11 + 5 2a = 16 Divide both sides by 2 a = 8

a = – ¾ 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: 9 - a + a = 12 + 3a + a What happens to the left-hand side? 9 – a + a = 12 + 3a + a 9 = 12 + 4a Take 12 from both sides 9 – 12 = 12 + 4a – 12 – 3 = 4a Divide both sides by 4 a = – ¾

Equations with fractions on both sides Example 15 Solve Multiply both sides by the LCD, 12. This kills the fractions. Put brackets around the numerators. 4 3 Cancel 4(3x + 1) = 3(2 – x) Expand 12x + 4 = 6 – 3x 15x = 2 x = 2/15

Equations with fractions on both sides Example 16 Solve Multiply both sides by the LCD, 20. This kills the fractions 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

Problem solving

A rectangular field is 5m longer than it is wide x Example 17 A rectangular field is 5m longer than it is wide x 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 So the width is 47.5m and Length is 47.5 + 5 = 52.5m The four sides total to 200………….. x + x + x + 5 + x + 5 = 200 4x + 10 = 200 Instead of writing x + 5 twice, you could have written 2(x + 5). This becomes 2x + 10 when you get rid of the brackets! Finally make sure they add to 200 4x = 190 x = 47.5

x + 12 Example 18 Another rectangular field is 12m longer than it is wide x Its perimeter is 1km. Find its dimensions (width and length). Let x equal the width. So the length is… x + 12 So the width is 244m and Length is 244 + 12 = 256m The four sides total to 1000………….. 4x + 24 = 1000 Finally make sure they add to 1000 4x = 976 x = 244

x = 17 Example 19 Find the value of x in this diagram As these are co-interior, they are supplementary and so must add to 180º 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! 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

Example 18 Find the value of b in this isosceles triangle As it’s isosceles, the other bottom angle must also be (2b + 1) The three angles add to 180, so….. (2b + 1)º 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! b – 7 + 2(2b + 1) = 180 b – 7 + 4b + 2 = 180 Expanding brackets 5b – 5 = 180 Cleaning up left side 5b = 185 Adding 5 to both sides b = 37 Dividing both sides by 5

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 So Jimmy has $4.15 and Joseph has………. 2x + 3.25 Mary has 2 x $4.15 = $8.30 Now we know they total to 24………….. x + 2x + 2x + 3.25 = 24 Joseph has $8.30 + $3.25 = $11.55 5x + 3.25 = 24 Finally make sure they add to $24 5x = 20.75 x = 4.15

Let the youngest (John) be x. Example 20 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 So John is 31 5x + 4 = 159 Mary is 2 x 31 = 62 5x = 155 Peter is 62 + 4 = 66 – 4 from both sides x = 31 divide both sides by 5

Example 21 x + 5 2x + 1 2x – 3 The isosceles triangle and the square have the same perimeter. Find x as a mixed numeral Triangle’s perimeter Square’s perimeter = x + 5 + 2(2x + 1) = 4(2x - 3) = x + 5 + 4x + 2 = 8x - 12 = 5x + 7 5x + 7 = 8x – 12 19 = 3x 5x + 7 – 5x = 8x – 12 – 5x 7 = 3x – 12 7 + 12 = 3x – 12 + 12

Example 22 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. x + x + x + 8 + x – 12 = 168 Sylvia is x – 12. 4x – 4 = 168 The twins are 43! 4x = 168 + 4 4x = 172 x = 43 Also, Marmaduke is 51, Sylvia is 31

Example 23 (3a – 5) 2(3a – 5) 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)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 Width = 49cm Length = 98cm 18a – 30 = 294 18a = 324 Area: 49 x 98 = 4802cm2 a = 18

Archie has 20, Muriel 30 and Ossie 45 Example 24 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 x + 1.5x + 2.5x – 5 = 95 So Muriel’s share is 1.5x 5x – 5 = 95 5x = 100 Oswald’s share is x + 1.5x – 5 = 2.5x – 5 x = 20 Archie has 20, Muriel 30 and Ossie 45

Example 25 5x - 4 3x + 2 20 – x The perimeter of this shape is 176 cm. Find the value of x And so is this side also 2x - 6 2x - 6 x + 14 This side is 5x – 4 – (3x + 2) So this side has to be 20 – x + 2x – 6 = 5x – 4 – 3x – 2 = 2x - 6 = x + 14 Now add up all the sides! 12x + 20 = 176 X = 13 12x = 156

Example 26 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?

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

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 Attila is 65 Peregrine is 52 Ugly is 72 Otto is 91 1.2x = 78 x = 65

Coins!! The solution 29 – x . Example 27 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? The solution 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 .

x coins each valued at 10c, and… (29 – x) coins each valued at 20c so we now have that there are…. x coins each valued at 10c, and… (29 – x) coins each valued at 20c 10x The x coins each valued at 10 cents must be worth a total of 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 Expand  10x + 580 – 20x = 410 580 – 10x = 410 Clean up 

– 10x = 410 – 580 – 10x = – 170 x = 17 Number of 10c = 17 Clean up  x = 17 Divide by – 10  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! Number of 10c = 17 Number of 20c = 12 Finally, check that 17 x 10 + 12 x 20 = 410

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

r ¼ x 2πr r r r r Factorising! Area = ¼ π r 2 + r 2 We set up an equation for the perimeter…. Left side + top + bottom + quadrant = 1000m r = 179.50755m r + r + 2r + = 1000m 4r + = 1000m Factorising! Area = ¼ π r 2 + r 2 Area = 57531 sq metres