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Algebra
Merit
Simplify
2x -12xy
2
6x
2
Simplify by factorising
2x - 12xy 2x ( x - 6y )
=
2
2
6x
6x
x - 6y
=
3x
2
Simplify by taking out the common
factor (2x) out of everything
2x -12xy x - 6y
=
2
6x
3x
2
Make W the subject
W
A=p
G
Make W the subject
W
A=p
G
2
2W
A =p
G
2
AG
=
W
2
p
Solve for x and y
2y + 3y = 15
-4x - 3y = 3
Solve for x and y
2y + 3y = 15 Þ 5y = 15 Þ y = 3
-4x - 3y = 3 Þ -4x - 9 = 3
-4x = 12 Þ x = -3
• A square warehouse is extended by 10 metres
at one end. The area of the extended
warehouse is 375m2
Find the original area of the warehouse.
Area = 152 =225 m2
x ( x + 10 )is= extended
375
• A square warehouse
by 10 metres
at one end. The
the= extended
x 2 +area
10x -of375
0
2
warehouse is (375m
x + 25 ) ( x - 15 ) = 0
Find the original
area
of
the
warehouse.
x = 15, x ¹ -25
Simplify
x - 6x - 16
x+2
2
Simplify
x - 6x -16 ( x - 8 ) ( x + 2 )
=
x+2
x+2
= x-8
2
• Elton has more than twice as many CDs as
Robbie. Altogether they have 56 CDs. Write a
relevant equation and use it find the least
number of CDs that Elton could have.
• Elton has more than twice as many CDs as
Robbie. Altogether they have 56 CDs. Write a
relevant equation and use it find the least
number of CDs that Elton could have.
R + 2R = 56
3R = 56
R = 18.666 Þ R = 18
E = 56 - 18 = 38
• Elton purchases some DVDs from the mall. He
buys four times as many music DVDs as movie
DVDs. The music DVDs are $2.50 each. The
movie DVDs are $1.50 each. Altogether he
spends $92.
Solve the equations to find out how many
music DVDs that he purchased.
S = 4V
2.5S +1.5V = 92
= 4Vpurchases some DVDs from the mall. He
•SElton
buys four times as many music DVDs as movie
2.5
4V
+
1.5V
=
92
DVDs. The music DVDs are $2.50 each. The
movie
DVDs
are
$1.50
each.
Altogether
he
10V + 1.5V = 92
spends $92.
Solve the
equations to find out how many
11.5V
= 92
music DVDs that he purchased.
(
)
V ( movie) = 8, S ( music ) = 32
Simplify
x x
+
2 8
Simplify
x x 4x + x 5x
+ =
=
2 8
8
8
One of the solutions of 4x2 + 8x + 3 = 0
is x = -1.5
• Use this solution to find the second solution of
the equation.
One of the solutions of 4x2 + 8x + 3 = 0
is x = -1.5
• Use this solution to find the second solution of
the equation.
( 2x + 3)
• Must be one of the brackets
One of the solutions of 4x2 + 8x + 3 = 0
is x = -1.5
• Use this solution to find the second solution of
the equation.
( 2x + 3)( 2x +1) = 0
• We need 2x to make 4x2
• We need +1 to make ‘3’
One of the solutions of 4x2 + 8x + 3 = 0
is x = -1.5
• Use this solution to find the second solution of
the equation.
( 2x + 3) ( 2x + 1) = 0
2
• We need
2x
to
make
4x
-3 -1
x =need +1, to make ‘3’
• We
2
2
The volume of the box shown is 60
litres. Find the dimensions of the box.
60 litres = 60, 000 cm3
V = 60000 = 50w ( w +10 )
60 litres = 60, 000 cm3
V = 60000 = 50w ( w + 10 )
w + 10w - 1200 = 0
2
( w + 40 ) ( w - 30 ) = 0
w = 30, w ¹ -40
Dimensions are 50cm:30cm:40cm
• The triangle drawn below is equilateral. The
perimeter is 30 cm. Write down two equations
and solve them simultaneously to find the
values of x and y.
•4y
The+triangle
below
2 = 2xdrawn
-yÞ
5yis-equilateral.
2x = -2 The
perimeter is 30 cm. Write down two equations
4y
2 = them
2y +simultaneously
x Þ 2y - x to
= find
-2 the
and+solve
values of x and y.
y = 2, x = 6
Simplify
x - 4y
2
x - 2xy
2
2
Factorise
x + 2y ) ( x - 2y )
x - 4y
(
=
2
x - 2xy
x ( x - 2y )
2
2
x + 2y
=
x
Express as a single fraction
x 3x
+
2 5
Express as a single fraction
x 3x 5x + 6x 11x
+
=
=
2 5
10
10
Solve the equation
x + 2x = 255
2
x + 2x = 255
2
Solve the equation
x + 2x - 255 = 0
2
( x + 17 ) ( x - 15 ) = 0
x = -17, 15
Simplify
2m m
+
3
4
Simplify
2m m 8m + 3m 11m
+ =
=
3
4
12
12
There are V litres in Claudia’s water tank. There are d “drippers”
on the irrigation hose from the tank to the garden. Each dripper
uses x litres of water per day.
• Write an expression to show the total amount
of water, T, left in the tank after one day.
There are V litres in Claudia’s water tank. There are d “drippers”
on the irrigation hose from the tank to the garden. Each dripper
uses x litres of water per day.
• Write an expression to show the total amount
of water, T, left in the tank after one day.
T = V - dx
There are V litres in Claudia’s water tank. There are d “drippers”
on the irrigation hose from the tank to the garden. Each dripper
uses x litres of water per day.
• At the end of the day on the 1st of April there
were 150 litres of water in the tank. The next
day, 4 drippers were used to irrigate the
garden and at the end of the day there were
60 litres of water left.
• Use the expression you gave above to show
how much water each dripper used on that
day.
T = V - dx
There are V litres in Claudia’s water tank. There are d “drippers”
on the irrigation hose from the tank to the garden. Each dripper
uses x litres of water per day.
T = V - dx
• At the end of the day on the 1st of April there
were 150 litres of water in the tank. The next
day, 4 drippers were used to irrigate the
garden and at the end of the day there were
60 litres of water left.
• Use the expression you gave above to show
how much water each dripper used on that
day.
60 = 150 - 4 x
4x = 90
x = 22.5 l / day
Graeme is designing a path around the front of
his garden. His design is shown below.
The width of the path is x metres.
Graeme has sufficient paving to make a path
with a total area of 22 m2.
• The area of the path
can be written as
• 4x+3x2 +(5-2x)x=22.
• Rewrite the equation
and then solve to find
the width of the path
around the front of the
garden.
The width of the path is x metres.
(
)
2
Graeme has sufficient
paving
4x + 3x + to5make
-2 2xa path
x=
with a total area of 22 m .
x + 9x - 22 = 0
2
( x + 11) ( x - 2 ) = 0
x = 2, x ¹ -11
The width of the path is x metres.
22