|
Factorisation
Clement
Radcliffe, Contributor
Today,
we will complete the review of factorisation
by going through the homework from
last week's lesson.
(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.
...
x2 + 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
- b2 = (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 is generally
preferred.
This
is illustrated as follows:
EXAMPLE
1
Solve
the simultaneous equations:
2x
- 2y = 1 (1)
7x
- 2y = 16 (2)
Subtracting
equation (2) from (1)
-5x
= -15
...
x = -15/-5 =3
Substituting
x = 3 into (1)
2
x 3 -2y = 1
...
6 - 2y = 6 -1 = 5
...
y = 5/2
...
y = 7
Answer
is x = 3, y = 5/2
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 -2 in both equations, you
eliminate y by subtracting? If the
coefficients differ in sign ONLY,
that is, if the coefficients of y
are -2 and +2, 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 use here.
Multiply
equation (2) x 3
...
-3x + 9y = -21 (3)
Add
equations (1) and (3)
...
7y = -14
...
y = -14/7 = -2
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)
3n + m = 2
4n
+ 3m = 3
{b)
2x - y = 1
3x
- y = 2
(c)
3x - 4y = 32
5x
+2y = 10
(d)
x + y = 7
2x + y = 10
(e)
2x = 11+ 3y
x + 2y + 12 = 0
(f)
3x + 2y = 1
4x - y = 16
Have
a good week.
|
|
|
Preparation
for the Caribbean Secondary
Education Certificate is ongoing
for these Anchovy High School
students.
-FILE
|
Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
|
