linear equation

Question

Divide 56 into two parts such that three times the first part exceeds one third of the second by 48.then the part is

Answer 1

x+y=56...y=56-x

3x=(y/3)+48...9x=y+144...y=9x-144

thus, 9x-144=56-x

10x=144+56

10x=200

x=20

y=56-x...=56-20...36

y=36

Answer 2

Let the two parts be x and 56 - x
3x = (1/3)(56 - x ) + 48
(10/3)x = 200/3
x = 20
Two parts are 20 and 36

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