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Answers
Step-by-step explanation:
Given :-
d is the HCF of 40 and 65.
To find :-
Find the value of x and y which satisfy
d = 40x+65y ?
Solution :-
Given that
d is the HCF of 40 and 65
40 = 2×2×2×5
65 = 5×13
HCF of 40 and 65 = 5
Therefore, d = 5
Now , Given condition d = 40x+65y
=> 5 = 40x+65y --------(A)
For finding the values of x and y we use Euclids Division Lemma
For any two Positive integers a and b there exist two positive integers q and r satisfying a = bq+r , where 0≤ r < b
Let a = 65 and b = 40
On applying Euclid's Division Lemma then
65 = 40×1 +25
=> 25 = 65-(1×40)-------(1)
Let a = 40 and b = 25
On applying Euclid's Division Lemma then
40 = 25×1 +15
=> 15 = 40-(1×25)------(2)
Let a = 25 and b = 15
On applying Euclid's Division Lemma then
25 = 15×1 +10
=> 10 = 25-(1×15) --------(3)
Let a = 15 and b = 10
On applying Euclid's Division Lemma then
15 = 10×1+5
=> 5 = 15-(1×10) ----------(4)
Let a = 10 and b = 5
On applying Euclid's Division Lemma then
10 = 5×2 + 0 -------------(5)
HCF (65,40) = 5
On taking (4)
15 = 10×1 + 5
It can be rearranged as
=> 5 = 15 - (10×1)
=> 5 = 15-1×10
From (3) we can write it as
=> 5 = 15 -1×(25-(1×15))
=> 5 = 15 -(25-1×15)
=> 5 = 15-(25-15)
=> 5 = 15-25+15
=> 5 = 2×15-25
From (2) we can write it as
=> 5 = 2×(40-1×25)-25
=> 5 = 2×(40-25)-25
=> 5 = 2×40-2×25-25
=> 5 = 2×40 -3×25
From (1) we can write it as
=> 5 = 2×40 -3×(65-1×40)
=> 5 = 2×40 -3×(65-40)
=> 5 = 2×40 -3×65+3×40
=> 5 = (2×40+3×40)-3×65
=> 5 = (2+3)×40 -3×65
=> 5 = 5×40 -3×65
=> 5 = (40×5)-(65×3)
=> 5 = (40×5)+(65×-3)---------(B)
We have , A and B are equal.
=> 40x+65y = (40×5)+(65×-3)
=> 40x + 65y = 40(5) + 65(-3)
On comparing both sides then
=> x = 5 and y = -3
Therefore, x = 5 and y = -3
Answer:-
The values of x and y for the given problem are 5 and -3 respectively.
Check:-
If x = 5 and y =-3 then 40x+65y
=> 40(5)+65(-3)
=> 200+(-195)
=> 200-195
=> 5
=> HCF of the two numbers 40 and 65
=> d = 5
Verified the given relations in the given problem
Used formulae:-
Euclid's Division Algorithm:-
For any two Positive integers a and b there exist two positive integers q and r satisfying a = bq+r , where 0≤ r < b
- Distributive Property under multiplication
a×(b+c) = (a×b)+(a×c)
Ex : 2×(40-25)-25
=> (2×40)-(2×25)-25
=> (2×40)-(3×25)
Points to know:-
- A Linear equation in two variables has infinitely number of many solutions.
- Euclid's Division Algorithm is used to find the HCF of two given numbers.