Question

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by2.

Answer 1

Let x and y be the two numbers.

2x + 3y = 92. -----eqn i

4x = 7y + 2
4x - 7y = 2. -----eqn ii

4 * eqn (i):
8x + 12y = 368. ---- eqn iii

2 * eqn (ii):
8x - 14y = 4. ---eqn iv

eqn (iii) - eqn (iv):
26y = 364

y = 364/26
= 14

Substituting in eqn (ii):
4x - 7(14) = 2
4x - 98 = 2
4x = 2 + 98
= 100
x = 100/4
= 25

Therefore, the numbers are:
x = 25 and y = 14

Answer 2

Answer:

Step-by-step explanation:

Given :-

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 Numbers

Solution :-

Let the first number be x and the second number be y.  

According to the Question

1st Equation

2x + 3y = 92 ……….(i)  

2nd Equation

4x - 7y = 2 ………(ii)  

On multiplying (i) by 7 and (ii) by 3,

14x + 21y = 644 ………..(iii)

12x - 21y = 6 ………..(iv)  

On adding (iii) and (iv), we get  

⇒ 26x = 650

⇒ x = 650/26

⇒ x = 25  

Putting the x values Eq in (i)  

⇒ 2x + 3y = 92

⇒ 2 × 25 + 3y = 92  

⇒ 50 + 3y = 92  

⇒ 3y = (92 – 50)

⇒ 3y = 42

⇒ y = 42/3 

⇒ y = 14

Hence, the first number is 25 and the second number is 14.