Question

dividend 56 into two parts such that three times the first part exceed one third of the second by 48 the parts are
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Answer 1

Answer:

20 and 36

Step-by-step explanation:

let the numbers be x and y

x + y = 56. equ (1)

3x = 1/3 y + 48

3x - y/3 = 48

(9x - y)/3 = 48

9x - y = 48 × 3

9x y = 144 equ (2)

add equ (1) + (2)

x + y + 9x - y = 56 + 144

10 x = 200

x = 200/10

x = 20

substitute x = 20 in equ (1)

x + y = 56

20 + y = 56

y = 56 - 20

y = 36

first number = 20

2nd number = 36

hope you get your answer