How do you solve the system a+2b=−2, −a+b+4c=−7, 2a+3b−c=5?

1 Answer
MathFact-orials.blogspot.com
Jan 5, 2017

(a,b,c)=(52,−27,18)

Explanation:

a+2b=−2
−a+b+4c=−7
2a+3b−c=5

I'm going to use the first equation to substitute for a in the other two equations:

a=−2b−2

~~~~~

−(−2b−2)+b+4c=−7

2b+2+b+4c=−7

3b+4c=−9

~~~~~

2(−2b−2)+3b−c=5

−4b−4+3b−c=5

−b−c=9

And we can now use this last equation to substitute into the one prior:

b=−c−9

~~~~~

3(−c−9)+4c=−9

−3c−27+4c=−9

c=18

And now let's find b:

−b−c=9

−b−18=9

b=−27

And check it:

3b+4c=−9

3b+4(18)=−9

3b+72=−9

3b=−81

b=−27

And now let's find the last variable:

a+2b=−2

a+2(−27)=−2

a−54=−2

a=52

and check it:

−a+b+4c=−7

−52−27+4(18)=−7

−79+72=−7

−7=−7

To summarize,

(a,b,c)=(52,−27,18)

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