Question 994597
{{{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=(132-x-13)/x}}}
Solving:
{{{6=(132-x-13)/x}}}-->{{{6=(119-x)/x}}}-->{{{6x=119-x}}}-->{{{6x+x=119}}}-->{{{7x=119}}}-->{{{x=119/7}}}-->{{{highlight(x=17)}}} .
Then {{{132-x=132-17}}}-->{{{highlight(132-x=115)}}}
The two parts are {{{highlight(17)}}} and {{{highlight(115)}}} .