Question

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

Answer 1

Answer . . .

Numbers = 20 and 7

_____________________

Let the -

• first number = x
• second number = y

Sum of the twice the first number and thrice the second number is 61.

According to question,

$\implies$ 2(x) + 3(y) = 61

$\implies$ 2x + 3y = 61 ...(1)

Also, said in question that

Subtract 5 times the second number from four times the first number the answer is 45.

According to question,

$\implies$ 4(x) - 5(y) = 45

$\implies$ 4x - 5y = 45 ...(2)

Multiply (eq 1) with 2

→ 2x + 3y = 61 (×2)

→ 4x + 6y = 122

→ 4x = 122 - 6y ...(3)

Substitute value of 4x in (eq 2)

$\implies$ 122 - 6y - 5y = 45

$\implies$ - 11y = 122 - 45

$\implies$ - 11y = - 77

$\implies$ y = 7

Substitute value of y in (eq 1)

$\implies$ 2x + 3(7) = 61

$\implies$ 2x + 21 = 61

$\implies$ 2x = 61 - 21

$\implies$ 2x = 40

$\implies$ x = 20

•°• First number = x = 20 and Second number = y = 7

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Verification:-

From the above calculations x = 20 and y = 7

Substitute value of x and y in (eq 1)

→ 2(20) + 3(7) = 61

→ 40 + 21 = 61

→ 61 = 61

Similarly, substitute value of x and y in (eq 2)

→ 4(20) - 5(7) = 45

→ 80 - 35 = 45

→ 45 = 45

Answer 2

$\large\underline\mathfrak\blue{Answer-}$

Numbers are 20 and 7.

______________________

$\large\underline\mathfrak\blue{Explanation-}$

Let the numbers be p and q.

$\bold\pink{According\:to\:the\:question-}$

1st condition : The sum of the twice the first number and thrice the second number is 61.

$\implies$ 2p + 3q = 61 _______(i)

2nd condition : when we subtract 5 times the second number from four time the first number the answer is a 45.

$\implies$ 4p - 5q = 45 ________(ii)

___________________

Multiply equation (i) by 2 and equation (ii) by 1.

{ 2p + 3q = 61 } × 2

{ 4p - 5q = 45 } × 1

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$\bold\pink{By\:elimination\:method-}$

$\implies$ 4p + 6q = 122

$\implies$ 4p - 5q = 45

By subtracting, we get,

$\implies$ 11q = 77

$\implies$ q = $\dfrac{77}{11}$

$\implies$ q = 7

_________________

Now, put the value of q in eq (i)

$\implies$ 2p + 3(7) = 61

$\implies$ 2p + 21 = 61

$\implies$ 2p = 61 - 21

$\implies$ 2p = 40

$\implies$ p = $\dfrac{40}{2}$

$\implies$ p = 20

$\therefore$ Numbers are 20 and 7.

_________________

$\large\underline\bold\blue{Verification-}$

We get,

p = 20 and q = 7

Put the above values in equation (i).

$\implies$ 2p + 3q = 61

$\implies$ 2(20) + 3(7) = 61

$\implies$ 40 + 21 = 61

$\implies$ 61 = 61

Hence verified!