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Linear
equations
Clement Radcliffe,Contributor
Let
us begin this week's lesson by reviewing
the answers to last week's homework.
1.
Evaluate: (2r - 3)3
SOLUTION
(2r
- 3)3 = (2r - 3)(2r - 3)(2r
- 3) = (4r2 - 6r - 6r +
9)(2r - 3)
=
(4r2 - 12r + 9)(2r - 3)
= 8r3 - 24r2
+ 18r -12r2 + 36r - 27
=
8r3 - 36r2 +
54r - 27
2.
Expand the following:
(a)
(M + 3) (M - 4)
(b)
(t - 3) (t + 6)
SOLUTION
(a)
(M + 3) (M - 4) = M2 +
3M - 4M - 12 = M2 - M -
12
(b)
(t - 3) (t + 6) = t2 -
3t + 6t - 18 = t2 + 3t
- 18
3.
Evaluate: (-2p +1)( -3p - 6)
SOLUTION
(-2p
+1)( -3p - 6) = 6p2 -3p
+ 12p - 6 = 6p2 + 9p -
6
4.
Simplify ( 2y - 1 ) /5 - ( y + 3 )
/ 2
SOLUTION
(
2y - 1 ) /5 - ( y + 3 ) / 2 The LCM
of 5 and 2 is 10.
(
2(2y - 1) - 5(y + 3) ) /10
=
( 4y - 2 - 5y - 15 )/10 = ( - y -
17 )/10
We
will now continue with linear equations.
LINEAR
EQUATIONS
The
inclusion of the equal sign differentiates
an equation from an algebraic expression.
This point is commonly missed by students
who sometimes attempt to solve algebraic
expressions. Do not fall into this
trap.
The
following points should be noted:
- Equations
identify either the relationship
between variables or the value of
a variable
- The
value of the variable is maintained
by performing identical operations
on both sides of the equation
- The
methods of clearing brackets and
simplifying algebraic expressions
are usually required to find solution
of equations
- In
order to solve the equation, one
approach is to simplify each side
of the equation and then equate
both sides
The
above is illustrated by the following
example:
Example
1
Solve
x⁄2 +
x⁄4 =
6
Simplify
the left hand side:
x⁄2
+ x⁄4
= 3x⁄4
Equating
both sides:
3x⁄4
= 6
x
= 24x⁄3
= 8
Example
2
Solve
( 4x + 5 )/4 - ( 9 + 2x )/3 = 0
Considering
the left hand side, the LCM of 3 and
4 is 12.
(
3(4x + 5) - 4(9 + 2x) )/12
=
( 12x + 15 - 36 - 8x )/12
=
( 4x - 21 )/12
Equating
both sides:
(
4x - 21 )/12 = 0 (cross-multiplying)
4x
- 21 = 0
x
= 21x⁄4
Alternatively,
you may multiply all terms by the
LCM of the denominators.
(
4x + 5 )/4 - ( 9 + 2x )/3 = 0
Multiply
both sides by 12:
3(4x
+ 5) - 4(9 + 2x) = 0
12x
+ 15 - 36 - 8x = 0
4x
- 21 = 0
x
= 21x⁄4
We
will now continue algebra with the
topic factorisation.
Note
that an algebraic expression is factorised
when it is expressed as the product
of its simplest factors. The usual
methods are:
(a)
Common factor
(b)
Grouping
(c)
Factorising of quadratic expressions
(d)
Difference of two squares.
The
methods are adequately explained in
the textbooks and you should use them
to aid you as you revise for your
exams.
It
is important that you do the following
in all cases:
(a)
Bring each factor to its simplest
form, for example, a factor 16x +
8 should be expressed as 4(4x + 2).
(b)
Check your answers, if you have the
time, by expanding and comparing the
result with the original expression.
This
week we will review the first two
methods of factorisation mentioned
above.
EXAMPLES
OF COMMON FACTOR METHOD
1.
Factorise: 9x2 -12x
The
common factor method is used, as 3x
is the factor which is common to both
terms. Both terms are divided by 3x
for us to obtain the second factor.
Answer: 3x(3x - 4)
Please
note that by expanding the answer,
3x(3x - 4) = 9x2 - 12x,
the given expression
2.
Factorise: 15x2y -10xy3
Note
that the common factor to both terms
is 5xy
By dividing each term by 5xy,
Answer is 5xy(3x - 2y2)
EXAMPLES
OF GROUPING METHOD
3.
Factorise ax + ay + bx + by
Note
that a is the common factor of ax
+ ay and b the common factor of bx
+ by
ax
+ ay + bx + by = a(x + y) + b(x +
y)
Do
you realise that (x + y) is common
to both expressions?
a(x + y) + b(x + y) = (x + y)(a +
b)
This method could, therefore, be described
as repeated common factor method.
4.
Factorise 2ax - 6ay + bx - 3by
2a(x
- 3y) + b (x - 3y)
=
(x - 3y)(2a+ b)
Homework
1.
Solve x⁄4
+ 16 = 2x
2.
Solve ( 2x - 3 )/2 - ( x + 4 )/4 =
1
3.
Factorise: (a) 7x2 - 21x
(b)
axy - a2y
4.
Factorise: 3x - 8y - 4xy + 6
Clement
Radcliffe is an independent contributor.
Send questions and comments to kerry-ann.hepburn@gleanerjm.com
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