Find the solution of the equation system

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SUMMARY

The discussion focuses on solving a system of equations defined as follows: \( a = \frac{b+c+d}{3} \), \( b = \frac{a+c+d}{5} \), \( c = \frac{a+b+d}{7} \), and \( d = c + 5600 \). The variables \( a, b, c, d \) are constrained to be positive. The solution involves substituting the equations iteratively to isolate each variable and determine their values based on the given relationships.

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Albert1
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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 $a,b,c,d>0$
 
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My solution:

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

$c=2100,\,\,a=4200,\,\,b=2800,\,\,d=7700$
 
Albert said:
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 $a,b,c,d>0$
let $a+b+c+d=24x$
then :$a=6x,b=4x,c=3x,d=11x=3x+5600$
$\therefore x=700$
and $a=4200,b=2800,c=2100,d=7700$
 

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