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 by 2

Answer 1

Hi ,

Let us assume x , y are two numbers ,

according to the problem given ,

2x + 3y = 92 ---( 1 )

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

multiply equation ( 1 ) with 2 and subtract

from ( 2 ) , we get

- 13y = - 182

y = ( -182 )/ ( -13 )

y = 14

substitute y = 14 in equation ( 2 ) , we get

4x - 7 × 14 = 2

4x - 98 = 2

4x = 2 + 98

4x = 100

x = 100/4

x = 25

Therefore ,

Required two numbers are ,

x = 25 ,

y = 14

I hope this helps you.
: )

Answer 2

Step-by-step explanation:

assume 2 no as x and y

equations 1

2x + 3y = 92....

equations 2

4x - 7y = 2

by eleminating we get

multiply the eq 1 by 2

we get

4x + 6y = 184...(

3)

4x + 6y = 184

4x -7 y = 2

- + -

o +13 y = 182

so

y is equal to 14

putting value of x in eq 2 we get

4x -7×14 = 2

so

4x = 100

x = 25