Question

find two numbers such that the sum of twice the first and thrice the second is 92,and for time the first exceeds seven times the second by 2​

Answer 1

EXPLANATION.

Let x and y are the two number.

To find the two number.

According to the question,

Case = 1.

The sum of twice the first and thrice the

second is 92

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

Case = 2.

The four times first exceeds seven

times the second by 2.

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

From equation (1) and (2) we get,

=> multiply equation (1) by 7

=> multiply equation(2) by 3

we get,

=> 14x + 21y = 644

=> 12x - 21y = 6

we get,

=> 26x = 650

=> x = 25

put the value of x = 25 in equation (2)

we get,

=> 4(25) - 7y = 2

=> 100 - 7y = 2

=> -7y = -98

=> y = 14

Therefore,

Two number are = 25 and 14

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The sum of twice the first and thrice the second is 92.

• The first exceeds seven times the second by 2.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The two numbers.

$\bf{\underline{\underline\green{Solution:-}}}$

Let the first number be x

And second number be y

According to the 1st condition:-

The sum of twice the first and thrice the second is 92.

➝ 2x + 3y = 92.....(i)

According to the 2nd condition:-

Four times the first exceeds seven times the second by 2.

➝ 4x - 7y = 2....…(ii)

Multiplying equation (i) with 2

➝ 2(2x + 3y = 92)

➝ 4x + 6y = 184.......(iii)

Subtracting equation (ii) from (iii)

➝ 4x + 6y - (4x - 7y) = 184 - 2

➝ 4x + 6y - 4x + 7y = 182

➝ 13y = 182

➝ y = 14

Substituting y = 14 in equation (i)

➝ 2x + 3y = 92

➝ 2x + 3(14) = 92

➝ 2x + 42 = 92

➝ 2x = 92 - 42

➝ 2x = 50

➝ x = 25.

We have :- x = 25 and y = 14

$\underline{ \boxed{ \rm \therefore The \: numbers \: are \: 14 \: and \: 25}}$