Question

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

Answer 1

`(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)`