Fraction Problem

Let's say cayden had x dollar. So Jayden had (784-x) dollar from 1st information.

From 2nd information:

x-x/6=(784-x)-124,

or, x*11/6=660

or, x=360.

So, cayden had 360$ at first.

Let, Cayden had $x and Jayden had             $y.

According to the question,

                  x + y = 784          ...............(1)

and,         x - \({x \over 6}\frac{}{}\) = y - 124

                5x - 6y = -744        ................(2)

Solving equation (1) and (2)

eq(1) *6 + eq(2)

              6x + 6y = 4704

              5x - 6y = -744

         ...............................................

                   11x = 3960

                       x = 360

Therefore, Cayden had $360 at first.

Let the amount clayden has be X

Given : 

sum of money clayden and jayden have=$784

So x+ amount jayden has =$784

Amount jayden has=$784-x

Clayden spent \( {1 \over 6}\frac{}{}\) =\({1 \over 6}\frac{}{}\)* X

So amount left with clayden=x-\(x = {x \over 6}{}^{}\frac{}{}\)

Amount left with clayden =\( {5x \over 6}\frac{}{}\)

Jayden spends $124 

So amount left with jayden = $784-$124-x

So amount left with jayden=$660-x

Given :both have same amount left 

Hence - 660-x=\( {5x \over 6}\frac{}{}\)

660*6-6x=5x

660*6=11x

60*6=x

360=x

Amount clayden has =$360

c+j =784

left money 

\( {5c \over 6}\frac{}{}\) = j- 124

\( {5c \over 6}\frac{}{}\)=784-c-124

\( {11c \over 6}\frac{}{}\)=660

c=360 (ans)