Question

the sum of twice the first and thrice the second is 92 and four times the first exceeds seven times then second by 2,then find the numbers

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:Let the first number be x and the second number be y. Then, we have: 2x + 3y = 92 ……….(i)

4x - 7y = 2 ………(ii)

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

14x + 21y = 644 ………..(iii)

12x - 21y = 6 ………..(iv)

On adding (iii) and (iv), we get

26x = 650 ⇒ x = 25

On substituting x = 25 in (i), we get

2 × 25 + 3y = 92

⇒ 50 + 3y = 92

⇒ 3y = (92 – 50) = 42

⇒ y = 14

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