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

$\left(a , b , c\right) = \left(52 , - 27 , 18\right)$

#### Explanation:

$a + 2 b = - 2$
$- a + b + 4 c = - 7$
$2 a + 3 b - c = 5$

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

$a = - 2 b - 2$

~~~~~

$- \left(- 2 b - 2\right) + b + 4 c = - 7$

$2 b + 2 + b + 4 c = - 7$

$3 b + 4 c = - 9$

~~~~~

$2 \left(- 2 b - 2\right) + 3 b - c = 5$

$- 4 b - 4 + 3 b - c = 5$

$- b - c = 9$

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

$b = - c - 9$

~~~~~

$3 \left(- c - 9\right) + 4 c = - 9$

$- 3 c - 27 + 4 c = - 9$

$c = 18$

And now let's find b:

$- b - c = 9$

$- b - 18 = 9$

$b = - 27$

And check it:

$3 b + 4 c = - 9$

$3 b + 4 \left(18\right) = - 9$

$3 b + 72 = - 9$

$3 b = - 81$

$b = - 27$

And now let's find the last variable:

$a + 2 b = - 2$

$a + 2 \left(- 27\right) = - 2$

$a - 54 = - 2$

$a = 52$

and check it:

$- a + b + 4 c = - 7$

$- 52 - 27 + 4 \left(18\right) = - 7$

$- 79 + 72 = - 7$

$- 7 = - 7$

To summarize,

$\left(a , b , c\right) = \left(52 , - 27 , 18\right)$