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Factorisation
(cont'd)
ClementRadcliffe, Contributor
Today,
we will complete the review of factorisation
by going through the homework from
last week's lesson.
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These students do needle work
before stitching these garments
during a clothing and textile
class at Ascot High School,
recently. -Anthony Minott/Freelance
Photographer
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(a)
factorise: X2
+ 7x + 12
In
this case, we find two numbers, a
and b, such that a + b = 7 and a x
b = 12. The numbers are 4 and 3.
...
x? + 7x + 12 = (x + 4) (x + 3).
(b) Factorise: 3x2 - 7x -6 using quadratic
factors:
3x2
- 7x -6 = (3x + 2) (x - 3).
(c)
Factorise: 3x -8y - 4xy + 6 using
the grouping method:
3x
+ 6 -8y - 4xy
3(x
+ 2) - 4y(2 + x)
=
(x + 2) (3 - 4y)
(d) Factorise: 9a2 - b2 by using the
difference of two squares method
9a2
- b? = (3a - b)(3a + b)
(e) Factorise: x2 - y2 - 4x + 4y factorising
(x
- y)(x + y) - 4(x - y)
=
(x - y) (x + y - 4)
(f) factorise: 16x2 -1 By using the
difference of two squares method
16x2
-1 = (4x - 1)(4x + 1)
You must ensure that you are familiar
with the four methods demonstrated
above and know when to use each.
Now
let us review another topic, simultaneous
linear equations.
SIMULTANEOUS
LINEAR EQUATIONS
The
solution of the simultaneous equations
is the pair of x and y values which
satisfy both equations.
If
both equations are plotted on a graph,
it is the point of intersection of
both lines.
You
may use the elimination or substitution
method. However, the former method
is generally preferred.
This
is illustrated as follows:
EXAMPLE
1
Solve
the simultaneous equations:
2x
- y = -1 - (1)
3x
-y = 2 - (2)
Subtracting
equation (2) from (1)
-x
= -3 ... x = 3
Substituting
x = 3 into (1)
...
6 - y = -1
...
y = 7
Answer
is x = 3, y = 7
You
may substitute the values x = 3 and
y = 7 into both equations in order
to check your answer.
Do
you realise that in
EXAMPLE 1, since the coefficient
of y is -1 in both equations, you
eliminate y by subtracting? If the
coefficients differ in sign ONLY,
that is, if the coefficients of y
are -1 and +1, then you eliminate
by adding.
EXAMPLE
2
Solve
the simultaneous equations:
5x
+ 3y = 31 - (1)
2x
+ y = 12 - (2)
Multiply
equation (2) by 3 and then subtract
equation (1) from equation (3).
6x
+ 3y = 36 - (3)
5x
+ 3y = 31 - (1)
...
x = 5
Substituting
x = 5 in (2)
...10
+ y = 12 - (2)
...
y = 12 - 10 = 2.
Answer:
x = 5 and y = 2
The
following is an example of the substitution
method:
EXAMPLE 3
Solve
the simultaneous equations:
5x
+ 3y =31
2x
+y =12
5x + 3y = 31 - (1)
2x
+y =12 .- (2)
From
equation (2), y = 12 - 2x
Substituting
into (1)
...
5x + 3(12 - 2x) =31 Clearing the brackets
5x
+ 36 - 6x = 31
...
- x =31 - 36
...
-x = -5 or x = 5
Substituting into equation (2)
...10
+ y =12
...
y = 2
...
Answer is: x = 5 and y = 2
Let us try another example:
3x
- 2y = 7 - (1)
-
x + 3y = -7 - (2)
The
elimination method is the appropriate
one to be used here.
Multiply
equation (2) x 3
...
-3x + 9y = -21 - (3)
Add
equations (1) and (3)
...
7y = -14
...
y = -14 = -2
7
Substitute
y = -2 in equation (1)
... 3x + 4 = 7
...
3x =7 - 4 = 3
...
x = 1
Answer
x = 1, y = - 2
Please attempt to solve the following
simultaneous equations:
(a)
n+ m = 7 (b) 2x - 2y = 1
2n
+ m = 7 7x - 2y = 17
(c)
5x + y = 16 (d) x + y = 7
x
- 2y = 1 2x + y = 10
(e)
2x = 11+ 3y (f) 2x + 3y = -1
x
+ 2y + 12 = 0 5x - 2y = 18
Have
a good week.
Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
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