Question

The system ⎧⎩⎨a2+b2−c2=05a−b=ca+b+c=30 has exactly one solution. find the solution by using substitution. enter your answer as an ordered triple in the format (a, b, c), like this: (42, 53, 64)

Answer 1

(5,12,13)

Explanation:

You are given the system of three equations

`\left\{\begin{array}{l}a^2+b^2-c^2=0\\5a-b=c\\a+b+c=30\end{array}\right.`

The second equation states that

`c=5a-b`

Substitute it into the first and third equations:

`\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.`

From the last equation

`a=5`

Substitute it into the first equation

`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`

Then

`c=5a-b=5\cdot 5-12=25-12=13`

The solution is (5,12,13)