SOLUTION: Separate 132 into two parts such that if the larger is divided by the smaller. The quotient is 6 and the remainder is 13.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Separate 132 into two parts such that if the larger is divided by the smaller. The quotient is 6 and the remainder is 13.      Log On


   

Question 994597: Separate 132 into two parts such that if the larger is divided by the smaller. The quotient is 6 and the remainder is 13.
Found 3 solutions by KMST, josgarithmetic, macston:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= the smaller part
132-x= the larger part
If the larger part, 132-x , is divided by the smaller, x, the quotient is 6 and the remainder is 13
means that 132-x , minus the 13 remainder, divided by x is exactly 6 :
6=%28132-x-13%29%2Fx
Solving:
6=%28132-x-13%29%2Fx-->6=%28119-x%29%2Fx-->6x=119-x-->6x%2Bx=119-->7x=119-->x=119%2F7-->highlight%28x=17%29 .
Then 132-x=132-17-->highlight%28132-x=115%29
The two parts are highlight%2817%29 and highlight%28115%29 .

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
This is just the description transcribed into algebraic symbolism.
system%28x%2By=132%2Cy%2Fx=6%2B13%2Fx%29.

Solve for x and y.

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
L=larger part; S=smaller part
.
S+L=132
L=132-S
.
L=6S+13
132-S=6S+13
119=7S
17=S
L=132-S=132-17=115
ANSWER: The two parts are 17 and 115.
.
CHECK:
L=6S+13
115=6(17)+13
115=102+13
115=115