Plz solve it plz plz mathematics

divide 56 into parts such that 3 times the ist number exceeds 1/3 times the 2nd number by 48 find the numbers...

let one part be x and the other part be 56-x

3x = 1/3(56-x) + 48
9x= 56-x+144

10x=200
x =20

56-x= 36

Let one part of 56 be 'x'. Then the other will be (56-x).

3 times first number=3x

1/3 times the 2nd number=(1/3)(56-x)

By the given condition, 3 times first number exceeds 1/3 times the second by 48.

So,

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

9x = 56-x+144
10x=200
x=20

So,
56-x=56-20=36

So, the number is divided into 20 and 36.

Let x be the first number and y be the second number:
Condition(1): x + y = 56
Condition(2): 3x - (1/3)y = 48
multply equation(2) by 3: 9x -y = 144 add this equation to equation(1)
result: 10x = 200,
hence x = 20 and y = 36.
check ans. by substitution.

LET Ist NUMBER BE X

SECOND NUMBER = 56 - X

3X = 1/3 ( 56 -X ) +48

9X = 56 - X + 144

10X = 200

X = 20

ANSWER NUMBERS ARE 20 & 36

suppose x & y are that two nos. now by using both the conditions x+y=56; and 3x-y/3 = 48
now you can solve these equations (any way) to get values of x & y, that's it.

Go to class 5th of any school....