Find the solution of the equation system

[Albert1]

find the solution of the equation system:
$\begin{cases}
a=\dfrac{b+c+d}{3}---(1)\\
b=\dfrac{a+c+d}{5}---(2)\\
c=\dfrac{a+b+d}{7}---(3)\\
d=c+5600\,\, ---(4)\end{cases}$
here $a,b,c,d>0$

 

[anemone]

My solution:

Spoiler

Rewrite the given top three equations as

$3a=b+c+d\,\,\rightarrow 4a=a+b+c+d---(5)$

$5b=a+c+d\,\,\rightarrow 6b=a+b+c+d---(6)$

$7c=a+b+d\,\,\rightarrow 8c=a+b+c+d---(7)$

We get $a=2c$, and $3b=4c$

By substituting $a=2c$, $3b=4c$ and $d=c+5600$ into $3a=b+c+d$ yields

$3(2c)=\dfrac{4c}{3}+c+c+5600$ and this gives

$c=2100,\,\,a=4200,\,\,b=2800,\,\,d=7700$

 

[Albert1]

find the solution of the equation system:
$\begin{cases}
a=\dfrac{b+c+d}{3}---(1)\\
b=\dfrac{a+c+d}{5}---(2)\\
c=\dfrac{a+b+d}{7}---(3)\\
d=c+5600\,\, ---(4)\end{cases}$
here $a,b,c,d>0$

Spoiler

let $a+b+c+d=24x$
then :$a=6x,b=4x,c=3x,d=11x=3x+5600$
$\therefore x=700$
and $a=4200,b=2800,c=2100,d=7700$