How to find a,ba, b and cc from this system of linear equations?

Find aa, bb and cc from:
a+c=3b+a=2c+b=−1
\begin{align}
a+c &= 3 \\
b+a &= 2 \\
c+b &= -1 \\
\end{align}

I tried the following way, the answer comes wrong:
(a+c)+(b+a)=3+2(a+c) + (b+a) = 3+2

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Hint: first equation tells us c=3−ac=3-a. Now substitute that into the third.
– lulu
Oct 20 at 18:16

1

 

Hint: Add the first two equations (as you did), then subtract the third.
– Ethan Bolker
Oct 20 at 18:20

  

 

Thanks, solved!
– Raven
Oct 20 at 18:28

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2 Answers
2

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(a+c)+(c+b)+(b+a)=3+2−1(a+c)+(c+b)+(b+a)=3+2-1
2(a+b+c)=42(a+b+c)=4
a+b+c=2a+b+c=2
Then if a+b+c=2a+b+c=2 and a+c=3a+c=3 then b=−1b=-1

Then if a+b+c=2a+b+c=2 and a+b=2a+b=2 then c=0c=0

Then if a+b+c=2a+b+c=2 and b+c=−1b+c=-1 then a=3a=3

a+c=3…..(1)a+c=3…..(1)
b+a=2…..(2)b+a=2…..(2)
c+b=−1…..(3)c+b=-1…..(3)

Try your way and solve like this-

(a+c)+(b+a)=3+2(a+c)+(b+a)=3+2
2a+c+b=5…….(4)2a+c+b=5…….(4)
Now, subtract equation(3) from (4)
2a+c+b=5−c−b=12a=6 \begin{array}\ 2a+c+b=5 \\\;\;\;\;\;-c-b=1\\ \hline2a\;\;\;\;\;\;\;\;\;\;\;=6
\end{array}
2a=6⟹a=32a=6\;\implies a=3
Now, Put the value value of aa in equations(1) and (2)-
3+c=3⟹c=03+c=3\;\implies c=0
b+3=2⟹b=−1b+3=2\;\implies b=-1

I hope it’ll help.