Question

there is a certain pair of numbers such that the sum of twice the first number and thrice the second number is 61 however when we subtract 5 times the second number from four times the first number the answer is 45 find the numbers​

Answer 1

Answer:

Answer: Given :  two numbers are such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

 

To find : the two numbers

 

Let x and y be the two numbers required.

 

 

According to the question :

=> 2x +  3y = 92 ...............................(1)

=> 4x- 7y = 2.....................................(2)

 

multiply the first equation by 2 , and subtract eq (1) from eq (2) , we get

 4x + 6 y = 184

- (4x - 7y  = 2)

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13 y            = 182

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=> y = 14 and  x = 25 Answer  

 

y = 14  and

Answer 2

Answer:

• first number be = x

• second number be = y

$\bold{A.T.Q}$

⟹ 2(x) + 3(y) = 61

⟹ 2x + 3y = 61 ...(1)

Now,

⟹ 4(x) - 5(y) = 45

⟹ 4x - 5y = 45 ...(2)

Multiply equation 1 with equation 2

→ 2x + 3y = 61 (×2)

→ 4x + 6y = 122

→ 4x = 122 - 6y ...(3)

Substitute value of 4x in (equation 2)

⟹ 122 - 6y - 5y = 45

⟹ - 11y = 122 - 45

⟹ - 11y = - 77

⟹ y = 7

Substitute value of y in (equation 1)

⟹ 2x + 3(7) = 61

⟹ 2x + 21 = 61

⟹ 2x = 61 - 21

⟹ 2x = 40

⟹ x = 20

• $\therefore$ First number = x = 20 

•  Second number = y = 7