1. 6 x + 25 xy
2
can be written as:
6 × x × x + 25 × x × y
(−5)2yx + 2x2 + 4x2
6 x 3 + 25 x 2 y
x
And you may have noticed that x(6x + 25y) also works.

2. Evaluate the following expressions when s = 6 and t = −1.
(a) 2(s + t)
(b) s + t2
(c) st + 3t
(d) t(s + 3)

3. Evaluate the following expressions when s = 6 and t = −1.
(a) 2(s + t) = 2 × (6 + (−1)) = 2 × 5 = 10
(b) s + t2 = 6 + (−1)2 = 6 + 1 = 7
(c) st + 3t = 6 × (−1) + 3 × (−1) = −9
(d) t(s + 3) = (−1) × (6 + 3) = −9

4. Identify the terms in the expression
4 − 3k + h − 27 k
3
.

5. Identify the terms in the expression
4 − 3k + h − 27 k
3
The terms are
+4
−3k
+ h3
−27 k

6. True or false?
In the expression 4a − a + 3 − 2b
2
(a) The coefficient of b is 2.
(b) a2 has no coefficient.
(c) The coefficient of a2 is 1.
(d) The coefficient of a2 is −1.
(e) 3 is a constant term.
(f) 3 − 2b is a term.
(g) a2 is a term.
(h) 4a and − a2 are like terms.

8. Simplify the following expressions by collecting like terms.
(a) k + 3k
(b) 2m + m2 − m
(c) 6xy − 3x − 2x + y
(d) 2ab − a2b − 5ba
(e) 4a + 2A + 3 − 1
(f) 4pq − 3q + pq +3q

9. Simplify the following expressions by collecting like terms.
(a) k + 3k = 4k
(b) 2m + m2 − m = m + m2
(c) 6xy − 3x − 2x + y = 6xy − 5x + y
(d) 2ab − a2b − 5ba = −3ab − a2b
(e) 4a + 2A + 3 − 1 = 4a + 2A + 2
(f) 4pq − 3q + pq + 3q = 5pq

10. Which of the following expressions are equivalent?
(a) xy + 3x − y
(b) xy − y + 3x
(c) y + xy + 3x
(d) y − xy + 3x
(e) 3x + yx − y

11. Write the following terms in their simplest forms.
(a) a×2×3×b×a
(b) 2a × 4ab
(c) t10 × 3t5
(d) a × (−b) × (−b)
(e) (−2) × g × 3h × (−g) × 7

12. Write the following terms in their simplest forms.
(a) a × 2 × 3 × b × a = 6a2b
(b) 2a × 4ab = 8a2b
(c) t10 × 3t5 = 3t15
(d) a × (−b) × (−b) = ab2
(e) (−2) × g × 3h × (−g) × 7 = 42g2h

13. Identify the terms in the following expressions and then simplify.
(a) 5a × 2b − (−a) × 4 × b
(b) 2 × (−7P) × P + Q × 8 × P × (−2)

14. Identify the terms in the following expressions and then simplify.
(a) The terms are + 5a × 2b and − (−a) × 4 × b.
5a × 2b − (−a) × 4 × b
= 10ab + 4ab = 14ab
(b) The terms are + 2 × (−7P) × P and + Q × 8 × P × (−2).
2 × (−7P) × P + Q × 8 × P × (−2)
= −14P2 − 16PQ

16. Multiply out the brackets in the following expressions.
(a) 2a(3a − 6)
(b) 2(x + x2)y
(c) −2(4 − 2a)
(d) a(b + c) + b(a − c)
(e) 2a − (b − a)

17. Multiply out the brackets in the following expressions.
(a) 2a(3a − 6) = 6a2 − 12a
(b) 2(x + x2)y = 2xy + 2x2y
(c) −2(4 − 2a) = −8 + 4a (or 4a − 8)
(d) a(b + c) + b(a − c) = ab + ac + ab − bc
= 2ab + ac − bc
(e) 2a − (b − a) = 2a − b + a = 3a − b

18. True or false?
(a) k − 3k = 2k
(b) a × b2 × (−2a) = −2a2b2
(c) 4 − (2 − x) = 2 + x
(d) a(b + c) − b(a − c) − c(b + a) = 0

19. Expand the following algebraic fractions and then simplify
where possible.
2A + 6
(a)
2
(b) 5 p + p 2 q − 3 pq
p

20. Expand the following algebraic fractions and then simplify
where possible.
2A + 6 2A 6
(a) = + = A+3
2 2 2
(b) 5 p + p 2 q − 3 pq 5 p p 2 q 3 pq
= + − = 5 + pq − 3q
p p p p

21. Start with a
Start with n
number
Think of a number
Double it
Add 1
Multiply by 5
Subtract 3
Subtract ten times the
number you first thought of
And the answer is…

22. Start with a
Start with n
number
Think of a number 5 n
Double it 10 2n
Add 1 11 2n + 1
Multiply by 5 55 5(2n + 1) = 10n + 5
Subtract 3 52 10n + 2
Subtract ten times the
number you first thought of
2 (10n + 2) − 10n = 2
And the answer is… 2 2

23. Start with a
Start with n
number
Think of a number
Double it
Subtract 3
Multiply by 2
Add 10
Divide by 4
Subtract 1
And the answer is…

24. Start with a
Start with n
number
Think of a number −1 n
Double it −2 2n
Subtract 3 −5 2n − 3
Multiply by 2 −10 2(2n −3) = 4n − 6
Add 10 0 4n + 4
4n + 4 4n 4
Divide by 4 0 = + = n +1
4 4 4
Subtract 1 −1 n
And the answer is… −1 n

25. Prove that the product of an even number and a multiple of 3 is
always divisible by 6.
(Hint: Write the even number as 2a and the multiple of 3 as 3b,
where a and b are integers.)

26. Prove that the product of an even number and a multiple of 3 is
always divisible by 6.
(Hint: Write the even number as 2a and the multiple of 3 as 3b,
where a and b are integers.)
Let the even number be 2a and the multiple of 3 be 3b where
a and b are integers.
Their product is (2a) × (3b) = 6 × ab.
ab is the product of integers and so is also an integer.
6 × ab is a multiple of 6 and so is divisible by 6.

27. Do you think that the sum of an even number and a multiple of
3 is divisible by 6?
Find a counterexample, that is, find two numbers, an even
number and a multiple of 3, whose sum is not divisible by 6.

28. Do you think that the sum of an even number and a multiple of
3 is divisible by 6?
Find a counterexample, that is, find two numbers, an even
number and a multiple of 3, whose sum is not divisible by 6.
A counterexample could be 4 (an even number) and 9 (a
multiple of 3).
Their sum, 13, is not divisible by 6.

30. Is x = 5 a solution to these equations?
(a) x + 1 = 4
(b) 3x − 4 = 1 + 2x
(c) 2(x − 1) − (13 − x) = 0

31. Is x = 5 a solution to these equations?
(a) x + 1 = 4
When x = 5, LHS = 5 + 1 = 6 ≠ 4
So x = 5 is not a solution to the equation.
(b) 3x − 4 = 1 + 2x
When x = 5, LHS = 3 × 5 − 4 = 11
and RHS = 1 + 2 × 5 = 11 = LHS.
So x = 5 is a solution to the equation.
(c) 2(x − 1) − (13 − x) = 0
When x = 5,
LHS = 2 × (5 − 1) − (13 − 5) = 2 × 4 − 8 = 0 = RHS.
So x = 5 is a solution to the equation.

32. Show (without solving the equation) that x = −2 is a solution
to the equation 3 − 5x = 15 + x.
Now solve the equation.

33. Show (without solving the equation) that x = −2 is a solution
to the equation 3 − 5x = 15 + x.
Now solve the equation.
When x = −2, LHS = 3 − 5 × (−2) = 13 and RHS = 15 + (−2) = 13.
LHS = RHS so −2 is a solution to the equation.
The equation is: 3 − 5x = 15 + x
Add 5x: 3 = 15 + 6x
Swap sides: 15 + 6x = 3
Subtract 15: 6x = −12
Divide by 2: x = −2
The solution is x = −2

34. Solve the following equations.
(a) 2a = 22
(b) a + 3 = 8
x
(c) =1
3
(d) 2k = 7 + k
(e) y = 12 − 2y

35. (a) The equation is: 2a = 22
Divide by 2: a = 11
The solution is a = 11.
(Check: if a = 11, LHS = 2 × 11 = 22 = RHS; the solution is correct.)
(b) The equation is: a+3=8
Subtract 3: a=5
The solution is a = 5.
(Check: if x = 5, LHS = 5 + 3 = 8 = RHS; the solution is correct.)
x
(c) =
The1equation is:
3
Multiply by 3: x = 3
The solution is x = 3.
(Check: if x = 3, LHS = 3/3 = 1 = RHS; the solution is correct.)

36. (d) The equation is: 2k = 7 + k
Subtract k: k=7
The solution is k = 7.
(Check: if k = 7, LHS = 2 × 7 = 14 and RHS = 7 + 7 = 14, so LHS = RHS and the
solution is correct.)
(e) The equation is: y = 12 − 2y
Add 2y: 3y = 12
Divide by 3: y=4
The solution is y = 4.
(Check: if y = 4, LHS = 4 and RHS = 12 − 2 × 4 = 4, so LHS = RHS and the
solution is correct.)

37. Solve the following equations.
(a) B + 13 = 7 − 5B
(b) x − 1 = 2(x + 1)
x+3
(c) =1
6
y
(d) 3( y − 7) = 14 −
2
(e) a a −5
= +1
3 6

38. (a) The equation is: B + 13 = 7 − 5B
Add 5B: 6B + 13 = 7
Subtract 13: 6B = −6
Divide by 6: B = −1
The solution is B = −1.
(Check: if B = −1, LHS = −1 + 13 = 12 and RHS = 7 − 5 × (−1) = 12, so LHS =
RHS and the solution is correct.)
(b) The equation is: x − 1 = 2(x + 1)
Multiply out the bracket: x − 1 = 2x + 2
Subtract x: −1 = x + 2
Swap sides: x + 2 = −1
Subtract 2: x = −3
The solution is x = −3.
(Check: if x = −3, LHS = −3 − 1 = −4 and RHS = 2 × (−3 + 1) = −4,
so LHS = RHS and the solution is correct.)

39. (c) The equation is: x+3
=1
6
Multiply by 6: x+3=6
Subtract 3: x=3
The solution is x = 3.
3+3
(Check: if x = 3, LHS = = 1 = RHS; the solution is correct.)
6
y
(d) The equation is: 3( y − 7) = 14 −
2
Multiply by 2: 6(y − 7) = 28 − y
Multiply out the bracket: 6y − 42 = 28 − y
Add y: 7y − 42 = 28
Add 42: 7y = 70
Divide by 10: y = 10
The solution is y = 10.
10
(Check: if y = 10, LHS = 3 × (10 − 7) = 9 and RHS = 14 −2 = 9,
so LHS = RHS and the solution is correct.)

40. (e) a equation is:
The a − 5
= +1
3 6
Multiply by 6: 2a = a − 5 + 6
Tidy up:2a = a + 1
Subtract a: a=1
The solution is a = 1.
1 1− 5 −4 1
(Check: if x = 1, LHS = and RHS = +1 = +1 = ,
3 6 6 3
so LHS = RHS and the solution is correct.)

42. A table has been reduced in price by 25%. It now costs £270. Express this
information as an equation and then solve the equation to find how much the
table cost before the reduction.
How would your equation need to change if the price reduction was 20%?

43. A table has been reduced in price by 25%. It now costs £270. Express this
information as an equation and then solve the equation to find how much the
table cost before the reduction.
How would your equation need to change if the price reduction was 20%?
Let the original price of the table be £P.
The price has been reduced by 25%, so the new price is 75% of £P, or £0.75P.
The table now costs £270. So 0.75P = 270.
Divide by 0.75: 270
P= = 360
0.75
The cost of the table before the reduction was £360.
(Check: 75% of £360 = 0.75 × £360 = £270.)
If the reduction was 20%, the equation would become 0.8P = 270.

44. Sara is five times the age she was 28 years ago. How old is Sara?

45. Sara is five times the age she was 28 years ago. How old is Sara?
Let Sara’s age now be A.
We’re told that Sara’s age 28 years ago × 5 = her age now.
28 years ago, Sara’s age was A − 28.
So the equation is 5 × (A − 28) = A
Multiply out the bracket: 5A − 140 = A
Subtract A: 4A − 140 = 0
Add 140: 4A = 140
Divide by 4: A = 35
So Sara is 35 years old.
(Check: 28 years ago, Sara was 7 years old and 5 times 7 gives 35, her present age.)

46. In four years time, Tom will be twice the age he was 11 years ago. How old is
Tom?

47. In four years time, Tom will be twice the age he was 11 years ago. How old is
Tom?
Let Tom’s age now be T.
We’re told that in four years time, Tom will be twice the age he was 11 years ago.
Putting this in a table:
Time Tom’s age
Now T
In 4 years time T+4
11 years ago T − 11
Create an equation: T + 4 = 2 × (T − 11)
Multiply out the bracket: T + 4 = 2T − 22
Subtract T: 4 = T − 22
Swap sides: T − 22 = 4
Add 22: T = 26
So Tom is 26 years old now.
(Check: in 4 years time Tom will be 30. 11 years ago, he was 15, and twice 15
gives 30.)

48. A man is five times as old as his son. In two years time, he will only be four
times as old as his son. How old is the son now?

49. A man is five times as old as his son. In two years time, he will only be four
times as old as his son. How old is the son now?
Let the son’s age now be s. The father is five times as old as his son, so he is 5s.
We’re told that in two years time, he will only be four times as old as his son.
Putting this in a table:
Time Son’s age Father’s age
Now s 5s
In 2 years time s+2 5s + 2
Create an equation: 5s + 2 = 4 × (s + 2)
Multiply out the bracket: 5s + 2 = 4s + 8
Subtract 4s: s+2=8
Subtract 2: s=6
So the son is 6 years old now.
(Check: if the son is 6, then his father is five times as old, 30. In two years time,
the son will be 8 and the father will be 32, which is 4 times the son’s age.)

50. Bob has 2 children, one two years older than the other. At present Bob is twice
as old as the sum of his children’s ages. In 14 years time, Bob’s age will be
exactly the sum of the ages of his two children. How old are the children now?

51. Bob has 2 children, one two years older than the other. At present Bob is twice as
old as the sum of his children’s ages. In 14 years time, Bob’s age will be exactly the
sum of the ages of his two children. How old are the children now?
Let the younger child’s age be x. Then the older child’s age is x + 2. Bob is twice as
old as the sum of their ages, so Bob’s age is 2 × (x + x + 2) = 2(2x + 2).
We’re told that in 14 years time, Bob’s age will be exactly the sum of the ages of his
two children. Putting this in a table:
Time Younger child’s age Older child’s age Bob’s age
Now x x+2 2(2x + 2)
In 14 years time x + 14 x + 16 2(2x + 2) + 14
Create an equation: 2(2x + 2) + 14 = (x + 14) + (x + 16)
Remove brackets: 4x + 4 + 14 = 2x + 30
Tidy up: 4x + 18 = 2x + 30
Subtract 2x: 2x + 18 = 30
Subtract 18: 2x = 12
Divide by 2: x=6
So the children are 6 and 8 years old.
(Check: if the children are 6 and 8, then Bob is now twice, the sum of their ages, which is 28.
In 14 years time, the children will be 20 and 22 and Bob will be 42, the sum of their ages.)

52. A number is divided by 7 and then 6 is added to the result giving 15. What
number did you start with?

53. A number is divided by 7 and then 6 is added to the result giving 15. What
number did you start with?
Let the number you start with be n.
Create an equation:
n
+ 6 = 15
7
Multiply by 7: n + 42 = 105
Subtract 42: n = 63
You start with the number 63.
(Check: if the number is 63, divide it by 7 to get 9, the add 6 to get 15.)

54. The sum of three consecutive integers is 351. Find the numbers.

55. The sum of three consecutive integers is 351. Find the numbers.
Let the middle one of the three integers be n.
Then the other two will be n − 1 and n + 1.
Adding these gives the equation:
(n − 1) + n + (n + 1) = 351
Tidy up the equation: 3n = 351
Divide by 3: n = 117
The three integers are 116, 117 and 118.
(Check: 116 + 117 + 118 = 351, as expected.)

56. Nicholas buys two books for his nieces, Amy and Becky, for Christmas.
Becky’s book costs £5 more than Amy’s and together the bill came to £21. How
much were the books?

57. Nicholas buys two books for his nieces, Amy and Becky, for Christmas.
Becky’s book costs £5 more than Amy’s and together the bill came to £21. How
much were the books?
Let the price of Amy’s book be £P.
Then Becky’s book cost five pounds more or £(P + 5).
The total cost was £21, giving the equation:
P + (P + 5) = 21
Tidy up: 2P + 5 = 21
Subtract 5: 2P = 16
Divide by 2: P=8
So the books were £8 and £13.
(Check: if the books are £8 and £13, then the total cost is £21, as expected.)

58. Great Uncle Theo is planning some donations to charity. He has £1000 which he
wants to divide between the Cats’ Home, the Dogs’ Home and the sanctuary for
parrots. He wants to give the Dogs’ Home twice as much as the Cats’ Home. He
also wants to give £100 more to the sanctuary for parrots than to the Cats’
Home. How much does he give to each?

59. Great Uncle Theo is planning some donations to charity. He has £1000 which he
wants to divide between the Cats’ Home, the Dogs’ Home and the sanctuary for
parrots. He wants to give the Dogs’ Home twice as much as the Cats’ Home. He
also wants to give £100 more to the sanctuary for parrots than to the Cats’
Home. How much does he give to each?
Let the amount that Great Uncle Theo gives to the Cats’ Home be £c.
Then he will give the Dogs’ Home twice this or £(2c), and the sanctuary
for parrots will get £(c + 100).
The total of the gifts is £1000. So we can create the equation:
c + 2c + (c + 100) = 1000
Tidy up: 4c + 100 = 1000
Subtract 100: 4c = 900
Divide by 4: c = 225
So Great Uncle Theo gives £225 to the Cats’ Home, £450 to the Dogs’
Home and £325 to the sanctuary for parrots.
(Check: £225 + £450 + £325 = £1000, as expected.)

60. There are four people travelling in a car. Bill and Ben are the same age, Molly is
twice as old as Bill and Pete is twenty years older than Molly. The sum of their
ages is 86. How old is Molly?

61. There are four people travelling in a car. Bill and Ben are the same age, Molly is
twice as old as Bill and Pete is twenty years older than Molly. The sum of their
ages is 86. How old is Molly?
Although you are asked to find Molly’s age, you may find it easier to start by
letting Bill’s age be b years. Then you will avoid fractions!
If Bill’s age is b years, then Ben’s age is also b years.
Molly is twice as old as Bill so she is 2b years old.
Pete is 20 years older than Molly so he is (2b + 20) years old.
The sum of their ages is 86, so we can create the equation:
b + b + 2b + (2b + 20) = 86
Tidy up: 6b +20 = 86
Subtract 20: 6b = 66
Divide by 6: b = 11
So Bill is 11 years old and from this we can work out that Molly is 22 and Pete is
42.
(Check: Adding the ages together, 11 + 11 + 22 + 42 = 86, as expected.)

62. Pirate Jack has left you instructions for finding the treasure.
‘Start at the big oak tree. Take a certain number of paces to the north, then take
that number plus 5 to the east. Finally, take half of the number of paces you took
to the north, but this time, go in a north easterly direction. There you will find
the treasure. In total you will take three times the number of paces you took to
the north.’
How many paces should you take to the north?
http://office.microsoft.com/en-gb/clipart

63. Pirate Jack has left you instructions for finding the treasure.
‘Start at the big oak tree. Take a certain number of paces to the north, then take that
number plus 5 to the east. Finally, take half of the number of paces you took to the north,
but this time, go in a north easterly direction. There you will find the treasure. In total
you will take three times the number of paces you took to the north.’ How many paces
should you take to the north?
Let the number of paces that you take to the north be x.
x
Then you must take (x + 5) paces to the east and paces to the north east.
2
The total number of paces is 3x.
x
Create an equation: x + ( x + 5) + = 3x
2
Tidy up: x
2 x + 5 + = 3x
2
Multiply by 2: 4x + 10 + x = 6x
Tidy up: 5x + 10 = 6x
Subtract 5x: 10 = x
Swap sides: x = 10
So you should take 10 paces to the north.
(Check: You take 10 paces to the north, then that number plus 5, which gives 15, to the east
and half the number you take to the north, which is 5, to the north east. In total, you take 10
+ 15 + 5 = 30 paces which is 3 times the number of paces you took to the north.)

64. A magic square has the property that the numbers in each column, each row and
each diagonal add to the same total.
Substitute a = 5, b = 3 and c = −1 in the square below and check that the result is
a magic square.
a+c a+b−c a−b
a−b−c a a+b+c
a+b a−b+c a−c
Now use algebra to explain why the property works for all values of a, b and c.

65. Substituting a = 5, b = 3 and c = −1 gives:
4 9 2
3 5 7
8 1 6
If you use algebra, you will find that the sum of each row, each column and
each diagonal is 3a. For example, the sum of the first row is
(a + c) + ( a + b − c) + (a − b) = 3a

66. 1. In the expression (4 − a) the coefficient of a is −1.
2. x + xy + x 2 = 3x + xy
3. 3(2x − 3) = 6x − 3
4. (2a + b) − (a + b) = a
5. x = 2 is a solution of the equation 2x + 1 = 5.
6. x = 1 is a solution of the equation 2x + 1 = 7 − x.