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Answers

Answered by tennetiraj86
2

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.
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