Answer:
(5,12,13)
Step-by-step explanation:
You are given the system of three equations
[tex]\left\{\begin{array}{l}a^2+b^2-c^2=0\\5a-b=c\\a+b+c=30\end{array}\right.[/tex]
The second equation states that
[tex]c=5a-b[/tex]
Substitute it into the first and third equations:
[tex]\left\{\begin{array}{l}a^2+b^2-(5a-b)^2=0\\a+b+5a-b=30\end{array}\right.\Rightarrow \left\{\begin{array}{l}a^2+b^2-(5a-b)^2=0\\6a=30\end{array}\right.[/tex]
From the last equation
[tex]a=5[/tex]
Substitute it into the first equation
[tex]5^2+b^2-(5\cdot 5-b)^2=0\\ \\25+b^2-(25-b)^2=0\\ \\25+b^2-625+50b-b^2=0\\ \\50b-600=0\\ \\50b=600\\ \\b=12[/tex]
Then
[tex]c=5a-b=5\cdot 5-12=25-12=13[/tex]
The solution is (5,12,13)
