Question

divide 57 into two parts such that one third of the first one seventh of the second together equal to 11 ​

Answer 1

Answer:

the two terms are 15 and 42

Step-by-step explanation:

Let x and y be the two terms

57=x+y_____eqn 1

1/3×x+1/7y=11

x/3+y/7=11

7x+3y=11×21

7x+3y=231____eqn 2

eqn 2 - 3(eqn 1)

7x+3y=231

3x+3y=171

---------

4x=60

x=15 substitute x=15 in eqn 1

you get y=42

so x= 15 and y=42

Answer 2

Answer:

15,42

Step-by-step explanation:

LET THE PARTS ARE X AND Y

THEN

x+y=57------------------(1)

sum of one third of the first one and seventh of the second

=x/3+y/7=11

multipying by LCM 21

7x+3y=231-----------------(2)

multiply (1) by 3 we get

3x+3y=171-------------------(3)

subtracting (3) from (2) we get

4x=60

x=15

then from (1)

15+y=57

y=57-15=42

so the parts of 57 are:

15 and 42