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Factorisation
Clement
Radcliffe, Contributor
 | Students
in performance at Denham Town High School's career day expo on Thursday, May 18.
- Ricardo Makyn/Staff Photographer |
Today,
we will complete the review of factorisation with the solution to the problems
given last week. (a)
Factorise: 3x -8y - 4xy + 6 Using the grouping method 3x
+ 6 -8y - 4xy
3(x + 2) - 4y(2 + x)
=
(x + 2) (3 - 4y)
Answer: (x + 2) (3 - 4y) (b)
Factorise: x²
- y² -
4x + 4y Factorising (x
- y)(x + y) - 4(x - y) =
(x - y) (x + y - 4) Answer:
(x - y) (x + y - 4) (c)
Factorise: 3x²
- 7x -6 =
(3x + 2)(x -3) Answer:
(3x + 2)(x -3) (d)
Factorise: 16x²
-1 By using the difference of two squares method 16x²
-1 = (4x - 1)(4x + 1) Answer:
(4x - 1)(4x + 1) (e)
Factorise: 9a²
- b².
By using the difference of two squares method 9a²
- b² =
(3a - b)(3a + b) Answer:
(3a - b)(3a + b) 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.
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) ie.
6 - y = -1 ie.
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) ie.
x = 5 Substituting
x = 5 in (2) ie.
10 + y = 12 ........ (2) ie.
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) ie.
5x + 3(12 - 2x) =31 Clearing the brackets 5x
+ 36 - 6x = 31 ie.
- x =31 - 36 ie.
-x = -5 or x = 5
Substituting
into equation (2)
ie.
10 + y =12 ie.
y = 2 ie.
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 ie.
-3x + 9y = -21 ........... (3) Add
equations (1) and (3) ie.
7y = -14 ie.
y = -14/7 = -2 Substitute
y = -2 in equation (1) ie.
3x + 4 = 7 ie.
3x =7 - 4 = 3 ie.
x = 1 Answer
x = 1, y = - 2. Please
attempt to solve the following simultaneous equations: a)2x
- 2y = 1
7x = 2y = 17
(b)x + y = 7
2x + y = 10
c)2x
= 11+ 3y
x
+ 2y + 12 = 0 Clement
Radcliffe teaches at Glenmuir High School. | 
