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Simultaneous
linear equations
Clement Radcliffe,Contributor
This
week 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:
The
terms are rearranged as follows:
3x
+ 6 -8y - 4xy
3(x
+ 2) - 4y(2 + x)
=
(x + 2) (3 - 4y)
(d)
Factorise: 9a2 - b2
The
difference of two squares method is
used.
As
a reminder, we must find the square
root of 9a2 and b2.
Since
the square roots are 3a and b, respectively,
then:
9a2
- b2 = (3a - b)(3a + b)
(e)
Factorise: x2 - y2
- 4x + 4y
Factorising
using the grouping method:
(x
- y)(x + y) - 4(x - y)
=
(x - y) (x + y - 4)
(f)
Factorise: 36x2 -1 By using
the difference of two squares method
36x2
-1 = (6x - 1)(6x + 1)
=
(6x - 1)(6x + 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
- 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 = 1
-2
y = 1 - 6 = - 5
y
= -5⁄-2
y
= 5⁄2
Answer
is x = 3, y = 5⁄2
You
may substitute the values x = 3 and
y = 5 into both equations in order
to check your answer. 2
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 be used 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
+ -2 x -2 = 7
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 (d) x + y = 7
4n
+ 3m = 3 2 x + y = 10
(b) 2x - y = 1 (e) 2x = 11 + 3y
3x
- y = 2 x + 2y + 12 = 0
(c) 3x - 4y = 32 (f) 3x + 2y = 1
5x
+ 2y = 10 4x - y = 16
Have
a good week.
Clement
Radcliffe is an independent contributor.
Send questions and comments to kerry-ann.hepburn@gleanerjm.com
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