Solving equations can be difficult. Have a look at this slideshow and see what can be done

1. Solving Equations
Russell Shaw
GCSE Mathematics
February 2010

2. Solving Equations
• Simple Equations
a + 6 = 10
• Visually
a =4
• Or Mathematically
a + 6 = 10
a = 10 – 6
a=4

3. Solving Equations
• Balancing Method
– Keep the balance level
– What you do to one side, do to the other
• Mathematically:
a + 6 = 10
a + 6 – 6 = 10 – 6
a=4

4. Solving Equations
• Opposite operation
– If its “add” on one side carry it over and “subtract”
– If its “subtract” on one side carry it over and “add”
– If its “multiply” on one side carry it over and
“divide”
– If its “divide” on one side carry it over and
“multiply”

5. Solving Equations
• Try these:
y+3=5
11 = y – 10
2a = 6
p/5 = 3
5t = 20
b/4 = 12

6. Solving Equations
• Try these:
y+3=5 y=2
11 = y – 10 y = 21
2a = 6 a=3
p/5 = 3 p = 15
5t = 20 t=4
b/4 = 12 b = 48

7. Solving Equations
• Combining operations:
2y + 3 = 21
• Need to get “unknowns” on one side and known
on the other
2y = 21 -3 = 18
2y = 18
• Now get the unknown by itself
y = 18/2 = 9
• Check your answer
2 x 9 + 3 = 18 + 3 = 21 √

8. Solving Equations
• Try these:
2y + 5 = 15
11 = 2y – 9
2a - 4 = 6
p/5 + 4 = 14
5t + 5 = 20

9. Solving Equations
• Try these:
2y + 5 = 15 y=5
11 = 2y – 9 y = 10
2a - 4 = 6 a=5
p/5 + 4 = 14 p = 50
5t + 5 = 20 t=3

10. Solving Equations
• Brackets:
3(2p + 5) = 45
• Expand the brackets first
• Then solve as normal
3 x 2p + 3 x 5 = 45
6p + 15 = 45
6p = 30
p=5
Check!!

11. Solving Equations
• Letters on both sides:
4(2x – 3) = 2x + 6
• Expand the brackets first
8x – 12 = 2x + 6
• Get the “unknowns” on one side and known
on the other
6x = 18
• Then solve as normal
x=3

12. Solving Equations
• Try these:
2y + 3 = 5y – 6
5t + 7 = 3t + 10
2(a + 3) = 7
3(5r + 2) = 12r – 8
6(g + 7) = 3(4g + 2)

13. Solving Equations
• Try these:
2y + 3 = 5y – 6 y=3
5t + 7 = 3t + 10 t = 3/2
2(a + 3) = 7 a = 1/2
3(5r + 2) = 12r – 8 r = -14/3
6(g + 7) = 3(4g + 2) g=6

14. Solving Equations
• Word Formulae
pay = rate of pay x hours worked + bonus
If rate of pay = £7/hour and hours worked = 40 and a
bonus of £20 is given, how much is the pay?
pay = £7 x 40 + £20 = £280 + £20 = £300
• A word formulae uses words to represent
relationships between quantities

15. Solving Equations
• Algebraic Formulae
pay = rate of pay x hours worked + bonus
We could represent pay as P, rate of pay as R, hours worked as H
and bonus as B
So
P=RxH+B
P = RH + B
If R = £8/hour and H = 30 and B = £10, how much is the pay?
pay = £8 x 30 + £10 = £240 + £10 = £250
An algebraic formulae uses letters to represent relationships
between quantities

16. Solving Equations
• Try this:
– Write a word formula and an algebraic formula for
the area of a rectangle
– Use your formula to find the area of a rectangle
with length = 10 and width = 3

17. Solving Equations
• More complicated formulae
s = ½ a t2
Find s when t = 4 and a = 10
s = ½ x 10 x 42
s = ½ x 10 x 16
s = ½ x 160
s = 80

18. Solving Equations
• Rearranging to change the subject of a
formula
– You will need to be able to rearrange formulae.
H = (4t + 6)/s
– Make t the subject of the formula
H = (4t + 6)/s
H x s = (4t + 6)
4t + 6 = Hs
4t = Hs – 6
t = (Hs – 6)/4

19. Solving Equations
• Rearranging to change the subject of a
formula
• Once you can rearrange you can solve
problems
– If Perimeter = 30cm and rectangle has length 8 cm
and width y cm – find y.
• P = 2 x l + 2 x w = 2l + 2w
• So P – 2l = 2w
• So w = (P – 2l)/2
• w = (30 – 2 x 8) / 2 = 7 cm

20. Solving Equations
• Rearranging to change the subject of a
formula
• Try this:
F = (9 x C) + 32
5
What is the temperature in Farenheit when it is 30O C
Rearrange so that the formula says C = …
What is the temperature in Centigrade when it is 212O F

21. Solving Equations
• Rearranging to change the subject of a
formula
• Try this:
F = (9 x C) + 32
5
What is the temperature in Farenheit when it is 30O C
F = (9 x 30 / 5) + 32 = 86O F
arrange so that the formula says C = …
C = 5 x (F – 32) / 9
What is the temperature in Centigrade when it is 212O F
C = 5 x (212-32) / 9 = (5 x 180) / 9 = 5 x 20 = 100O C

22. Solving Equations
• Inequalities
• Values maybe more than… or less than…
• > means greater than
• < means less than
• ≥ means greater than or equal to
• ≤ means less than or equal to

23. Solving Equations
• Inequalities
• Things maybe more than… or less than…
• 6 is greater than 4 6 > 4
• 5 is less than 10 5 < 10
• x is greater than or equal to 5 x ≥ 5
• 2 is less than or equal to y 2≤y

24. Solving Equations
• Inequalities on a number line
• We can represent inequalities on a number line.
• If the number is “included” – using a ≥ or ≤ symbol – we
make the end point a solid circle
• x ≥ -1
• If the number is “excluded” – using a > or < symbol – we
make the end point an empty circle
• x<2

25. Solving Equations
• Inequalities on a number line
• We can represent dual inequalities on a number
line.
• -3 < x ≤ 4 (-3 is less than x, and x is less than or
equal to 4)
• -4 ≤ x ≤ 3

26. Solving Equations
• Inequalities on a number line
• Draw a number line from -10 to 10
• Try these:
x≥5
x<3
x < -5
x>0
-3 < x < 7
-9 ≤ x < 3

27. Solving Equations
• Solving Inequalities
• You maybe asked to give the integer solutions to
an inequality
• -3 < x ≤ 4
-3 is NOT included (<), BUT 4 is included (≤) so the
solution is: -2 , -1, 0, 1, 2, 3, 4

28. Solving Equations
• Solving Inequalities
• Find integer solutions to:
-2 ≤ n ≤ 1
0<n<5
• Write an inequality for the integers listed:
-2, -1 , 0 , 1, 2, 3, 4
1, 2, 3, 4, 5, 6

29. Solving Equations
• Solving Inequalities
• Find integer solutions to:
-2 ≤ n ≤ 1 -2, -1, 0, 1
0<n<5 1, 2, 3, 4
• Write an inequality for the integers listed:
-2, -1 , 0 , 1, 2, 3, 4 -2 ≤ n < 5 or -1 < n ≤ 4 or ..
1, 2, 3, 4, 5, 6 0 < n < 7 or 1 ≤ n < 7 or ..

30. Solving Equations (H)
• Solving Inequalities
• We can solve inequalities – just like equations
• Solve 3x + 3 > 18
3x > 18 – 3 = 15
x > 15/3 = 5
x >5
• These can have brackets included too –just like
normal equations!

31. Solving Equations (H)
• Solving Inequalities
• We can solve inequalities – just like equations
• When solving you can:
– Add or subtract same quantity to both sides
– Multiply or divide both sides by the same positive
quantity
• BUT you can’t:
– Multiply or divide both sides by a negative quantity

32. Solving Equations (H)
• Solving Inequalities
• You can solve two sided inequalities by treating it
as two separate inequalities.
• 7 ≤ 3x -2 < 10
So solve: 7 ≤ 3x -2 and 3x -2 < 10
Writing the answer as 3 ≤ x < 4

33. Solving Equations (H)
• Solving Inequalities
• Try these:
3p – 5 ≤ 4
1 + 6b > 7
8 – 2m ≥ 1 – 4m
7 < 4p + 3 ≤ 27
3 < 2q – 7 < 7

34. Solving Equations (H)
• Solving Inequalities
• Try these:
3p – 5 ≤ 4 p≤3
1 + 6b > 7 b>1
8 – 2m ≥ 1 – 4m m ≥ -7/2
7 < 4p + 3 ≤ 27 1<p≤6
3 < 2q – 7 < 7 5<q<7