Question

Please Solve!!!!!

If two numbers are such that the sum of four times the first and three times the second is 51 and 5 times the first exceeds seven times the second by 10. Find the numbers.

Answer 1

Let the nos be x and y.
4x+3y=51. --*by7
5x-7y=10. --*by3

28x+21y=357
15x-21y=30

43x=387
x=9
5*9-7y=10
45-7y=10
7y=35
y=5

Answer 2

Hey Friend
Here is your answer

Let first number = x
Let second number = y
4x + 3y = 51 -----(1)
5x = y + 10
x = y + 10 / 5 -----(2)
Put (2) in (1)
4 ( y + 10 / 5 ) + 3y = 51
4y + 40 / 5 + 3y = 51
4y + 40 + 15y / 5 = 51
40 + 55y = 51 * 5
55y = 255 - 40
y = 215 / 55
y = 3.90 --------(3)
Put (3) in (2)
x = 3.90 + 10 / 5
x = 13.90 / 5
x = 2.78

Hope this helped you
Have a nice day.