# How do you solve the system a + b + 2c = -11, a - 6b +3c = 22, 2a - 4b + c = 14?

May 2, 2016

$S = \left\{\left(- 2 , - 5 , - 2\right)\right\}$

#### Explanation:

Use Elimination Method, Matrices, Determinants, or Graphing to solve. I am going to use the Elimination Method.

$a + b + 2 c = - 11$----> Equation 1 (Eq.1)
$a - 6 b + 3 c = 22$-->Equation 2 (Eq.2)
$2 a - 4 b + c = 14$----> Equation 3 (Eq.3)

Multiply Eq.1 by -1 and then add to Eq.2
$\left[- a - b - 2 c = 11\right]$
+$\left[a - 6 b + 3 c = 22\right]$
$- 7 b + c = 33$----> Result 1 (R1)

Multiply Eq.2 by -2 and then add to Eq.3
$\left[- 2 a + 12 b - 6 c = - 44\right]$
+$\left[2 a - 4 b + c = 14\right]$
$8 b - 5 c = - 30$---> Result 2 (R2)

Multiply R1 by 5 and add to R2
$\left[- 35 b + 5 c = 165\right]$
+$8 b - 5 c = - 30$
$- 27 b = 135$
$b = - 5$

Put -5 in for b in R1
$- 7 \left(- 5\right) + c = 33$
$35 + c = 33$
$c = 33 - 35 = - 2$

Put -5 for b and -2 for c into Equation 1 and solve for a.
$a - 5 - 4 = - 11$
$a - 9 = - 11$
$a = - 11 + 9$
$a = - 2$

$S = \left\{\left(- 2 , - 5 , - 2\right)\right\}$