# DIVIDE 56 INTO TWO PARTS SUCH THAT THREE TIMES THE FIRST PART EXCEEDS ONE THIRD OF THE SECOND BY 48. THE PARTS ARE.

Assume two parts are x and y.

3x = y/3 + 48

9x -y=144...(1)

x +y = 56....(2)

Equation (1) + Equation (2)

10 x = 144 + 56

x= 200/10 = 20

y = 56 - 20 = 36

So such two number are 20 and 36.

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