SOLUTION: Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are ?

Question 1043819: Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are ?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
the first part is x
the second part is 56-x.
3x-48 (three times the first exceeds, so subtract 48 to make it equal)=(1/3)(56-x)
3x-48=(1/3)(56-x). Multiply both sides by 3 to remove the fraction
9x-144=56-x
10x=200
x=20, first part.
3 times that is 60
subtract 48 and it is 12
that is 1/3 the second, so 3 times that is the second, and that is 36, second part.
20 and 36 equal 56.

RELATED QUESTIONS
Divide 56 into two parts such that three times the first part exceeds one third of the... (answered by josgarithmetic)

Divide 56 into two parts such that three times the first part exceeds one third of the... (answered by math_helper)

Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The... (answered by lynnlo)

Divide 56 into two equal parts such that 3 times the first part exceeds one third of the... (answered by ikleyn)

Divided 56 into two parts such that three times the first part exceed one third of second (answered by stanbon)

Divide 20 into two parts such that three times the square of one part exceeds the other... (answered by amarjeeth123)

Divide 32 into two parts such that five times of the first part exceeds four times of the (answered by Maths68)

Divide 20 into two parts such thay three times the square of one part exceeds the other... (answered by ewatrrr)

Divide $128 into two parts such that one-third of the first part is less than 1 5 of... (answered by ewatrrr)