Question

divide 56into two parts such that three times the first part exceeds one third of the second by 48. the parts are​

Answer 1

Answer:

The parts are 20 and 36

Step-by-step explanation:

Let the parts be a & b

ATQ

a  = 56 - b     --------(1)

Also, 3a = b/3 + 48

Insert value from (1)

3(56-b) = b/3 + 144/3

9(56-b) = b + 144

504 - 9b = b + 144

504 - 144 = 10b

b = 360/10

b = 36

a = 56-36 = 20

Answer 2

Answer:

Let the first part be x therefore the second part will be 56-x.

3x-1/3(56-x) =48

(9x-56+x) =144

10x=200

x=20

56-x=36