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!

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!

9 + 6 = 15

Begin with a true statement

Add 3 to both sides9 + 6 + 3 = 15 + 318 = 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

x – 9 + 9 = 5 + 9 x = 14 x = 5 + 9

ZERO!

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

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

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:

123

y

We can now cancel the 3s on the left side, and at the same time work out 12 x 3

y = 36

x3x3

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

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

This trick is really useful in equations where there is a negative in front of the letter!

Solve – a + 7 = 12Change 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!

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

Solve 74

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 44

a= 7

Cancel the 4s on the left

Example 8:

Solve6

5

37

a

First we multiply by 5 to get rid of the fraction

65

37

ax 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 = – 30Add 7 to both sides

– 7 + 3a + 7 = – 30 + 7 3a = – 23

Divide both sides by 3 a = – 7.667

Example 9:

Solve6

5

34

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

25

3

a

4645

34

a– 4 – 4

Example 10:

Solve 47

29

a

Add 9 to both sides

Multiply both sides by 7

Divide both sides by 2

a = 45.5

137

2

a

Sign change 47

29

a

2a = 91

Solve 2(x – 5) = 15

Expand the brackets

2x – 10 = 15Add 10 to both sides

2x – 10 + 10 = 15 + 102x = 25

Divide both sides by 2 x = 25/2 or 12.5

Example 11……….

Solve 2(x – 5) + 3(2x + 1) = 28Expand the brackets

2x – 10 + 6x + 3 = 28

Add 7 to both sides

8x – 7 + 7 = 28 + 78x = 35

Divide both sides by 8 x = 35/8 or 4.375

Example 12……….

Clean up left side 8x – 7 = 28

Equations with an unknown on both sidesExample 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!

Equations with an unknown on both sidesExample 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 =

Equations with fractions on both sidesExample 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

Equations with fractions on both sidesExample 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

Example 17

A rectangular field is 5m

longer than it is wideIts 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 = 2004x + 10 = 200 4x = 190

x = 47.5

So the width is 47.5m

Length is 47.5 + 5 = 52.5m

Finally make surethey 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!

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 + 12and

The four sides total to 1000…………..

4x + 24 = 10004x = 976

x = 244

So the width is 244m

Length is 244 + 12 = 256m

Finally make surethey add to 1000

x

x + 12

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 = 180Clean up left side

11x – 7 = 180Add 7 to both sides 11x = 187Divide 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!

Example 18Find 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

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 = 245x = 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 surethey add to $24

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

The isosceles triangle and the square have the same perimeter. Find x as a mixed numeral

2x – 3

x + 5

2x +

1

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

16x

Example 21

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 = 1684x = 168 + 44x = 172 x = 43

The twins are 43!

Also, Marmaduke is 51, Sylvia is 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) = 29418a – 30 = 294

18a = 324 a = 18

Width = 49cmLength = 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)

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 shareSo Muriel’s share is 1.5xOswald’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

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

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

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 = xPeregrine’s age = 0.8xUgly’s age = Peregrine’s age + 20

= 0.8x + 20Otto’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 = 0.8xUgly = 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 = 78x = 65

Attila is 65Peregrine is 52Ugly is 72Otto is 91

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

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 10xand 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

– 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

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

rr

r

r

¼ x 2πr

We set up an equation for the perimeter….

Left side + top + bottom + quadrant = 1000m

r + r + 2r + = 1000mr2

4r + = 1000mr2

10002

4

r

24

1000

r

r = 179.50755m

Area = ¼ π r 2 + r

2 Area = 57531 sq metres

Factorising!