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2.5
HCF(GCD) and LCM of Algebraic terms:
In the case of numbers, we have understood what HCF (GCD)
of two or more numbers is. It is the highest among all common factors of the
numbers.
Let us take an example of 4 numbers 4, 8, 20, 16. We know
by inspection that 2 and 4 are common factors of all the four numbers and
highest among them is 4.
Therefore HCF of 4, 8, 20, and 16 is 4.
HCF is useful in simplifying
fractions.
Let us examine how we simplify the fraction 30/48.
First, we find HCF of 30 and 48 which happens to be 6.Then
we express both numerator and denominator as a product of this HCF.
(30 = 6*5 and 48 = 6*8)
Next, we cancel this common factor from both numerator and
denominator to get the proper fraction
(I.e. 30/48 = 6*5/6*8 = 5/8)
What is LCM? It is
the lowest number among all the common multiples of the numbers.
Let us take the same example of numbers 4, 8, 20, 16.
We notice that 80, 160, 320
are all common multiples of
the above numbers. The least among these common multiples is 80 and 80 is
called LCM
LCM is useful in adding fractions.
Let us add 1/4, 1/8, 1/20
LCM of 4,8,20 is 40
1/4 = 10/40
1/8 = 5/40
1/20 = 2/40
1/4+1/8+1/20 = 10/40+5/40+2/40 = (10+5+2)/40 = 17/40
For finding HCF and LCM of algebraic terms, we follow same
method that we use in the case of finding HCF and LCM of numbers.
Let us recollect Finding HCF and LCM of numbers.
Finding HCF
Step |
Procedure |
1 |
List
all the numbers |
2 |
On the
left side write the smallest common divisor for all of these numbers |
3 |
On the
2nd line write the quotients |
4 |
On the
left side write the common divisor for all of these quotients as got in the step2 |
5 |
Repeat
these steps till there are no more common factors for
all the numbers |
Product of the common divisors (appearing on the left hand
side) is HCF of the numbers
Find HCF of 16,24,20
2 | 16,24,20
2 | 8,12,10
4, 6, 5
We stop further divisions as the remaining numbers(all) ( 4 6 and 5) do
not have common factors other than 1
Therefore (2*2) = 4 is HCF of 16,24,20
Finding LCM
Step |
Procedure |
1 |
List
all the numbers |
2 |
On the
left side write the common divisor for any of
these numbers |
3 |
On the
2nd line write the quotients |
4 |
On the
left side write the common divisor for any of
these quotients as got in the step2 |
5 |
Repeat
these steps till there are no more common factors among any two numbers |
Product of the common divisors and the remaining numbers
on the last line is LCM
Find LCM of 16,24,20
2 | 16,24,20
2 | 8,12,10
2 | 4,6,5
| 2,3,5
We stop here further divisions as none
of the remaining numbers (2,3,5) have common factors other than 1
Thus LCM = (2*2*2)*(2*3*5) = 240
Also observe that, product of HCF and LCM of
2 numbers = Product of 2 numbers.
This relationship holds good for algebraic terms also,
hence if two terms and their HCF or LCM are given we can find their LCM or HCF
respectively.
2.5
Problem 1 : Find HCF of 16a4b3x3,
24b2m3n4y, 20a2b3nx3
Solution:
1.HCF of co-efficients (16,24,20) is 4
The variables are a4b3x3,
b2m3n4y, a2b3nx3
and we notice that b is a common factor of the variables
4b | 16a4b3x3,
24b2m3n4y, 20a2b3nx3
( We start dividing by 4b as it is the common factor for all the terms)
b | 4a4b2x3,
6bm3n4y, 5a2b2nx3 (b is a common factor of all
terms)
|4a4bx3, 6m3n4y,
5a2bnx3
We need to stop
further division as there are no more common factors among all the terms
4b*b= 4b2 is HCF of the term
HCF is useful for taking common factors out
in addition/subtractions and in simplifying expressions
Let us simplify 16a4b3x3+24b2m3n4y-
20a2b3nx3
16a4b3x3+24b2m3n4y-
20a2b3nx3
=4b2(4a4bx3+6m3n4y-
5a2bnx3)
2.5
Problem 2 : Find HCF and LCM of 6x2y3,
8x3y2, 12x4y3, 10x3y4
Solution:
1.HCF of co-efficients (6,8,12,10) is 2
The variables are x2y3, x3y2,
x4y3, x3y4and we notice that x is a
common factor of the variables
a)
Finding
HCF
2x | 6x2y3, 8x3y2,
12x4y3, 10x3y4 (2x is a common factor of all the terms)
x | 3xy3,
4x2y2, 6x3y3, 5x2y4
y |3y3,
4xy2, 6x2y3,
5xy4
y |3y2,
4xy, 6x2y2,
5xy3
3y, 4x, 6x2y, 5xy2
We need to stop further division as there are no more
common factors among all the terms
2x*x*y*y = 2x2y2 is HCF of variable
Use of HCF:
Let us simplify 6x2y3+8x3y2-12x4y3+10x3y4
6x2y3+8x3y2-12x4y3+10x3y4
= 2x2y2(3y+4x-6x2y+5xy2)
a)
Finding
LCM
2x | 6x2y3, 8x3y2,
12x4y3, 10x3y4 (2x is a common factor of all the terms)
x | 3xy3,
4x2y2, 6x3y3, 5x2y4
y |3y3,
4xy2, 6x2y3,
5xy4
y |3y2,
4xy, 6x2y2,
5xy3
Y
|3y, 4x, 6x2y , 5xy2
x | 3, 4x 6x2,
5xy
2
| 3, 4 6x
5y
ญญญญญญญญญญญญญญ 3
| 3, 2
3x 5y
| 1, 2 x 5y
We stop further division, as there are no more common
factors among any two terms.
Thus LCM = ( 2x*x*y*Y)*(Y*x*2*3*2*x*5y)
=2x2y2* 60x2y2 = 120x4y4
Use of LCM:
Simplify (1/6x2y3)+(1/8x3y2)-(1/12x4y3
)+(1/10x3y4)
Note
(1/6x2y3) = (20x2y/120x4y4)
(1/8x3y2) = (15xy2/120x4y4)
(1/12x4y3) = (10y/120x4y4)
(1/10x3y4) = (12x/120x4y4)
(1/6x2y3)+(1/8x3y2)-(1/12x4y3
)+(1/10x3y4)
= (20x2y+15xy2-10y+12x)๗(120x4y4)
2.5 Summary of learning
No |
Points studied |
1 |
Finding
HCF and LCM of algebraic terms by division method |