How do you solve the system x+y+z=1, -2x+2y+3z=20, and 2x-2y-z=-16?

Sep 28, 2016

$x = - 4 , y = 3 , z = 2$

Explanation:

The 2 main ways of solving a system of equations are elimination and substitution.

The ideal scenario for elimination is if there are additive inverses in two of the equations.. for instance: $5 x - 5 x = 0 , - 3 y + 3 y = 0$

We have 3 equations with 3 unknowns.

Notice that in equations B and C there are 2 pairs of additive inverses!
$\textcolor{w h i t e}{\times \times \times \times} x + \text{y+""z=" } 1$......................A

$\textcolor{w h i t e}{\times \times \times} \textcolor{red}{- 2 x} \textcolor{b l u e}{+ 2 y} + 3 z = \text{ } 20$..................B
$\textcolor{w h i t e}{\times \times \frac{\times}{.} x} \textcolor{red}{2 x} \textcolor{b l u e}{- 2 y} - z = - 16$..................C
$B + C : \textcolor{w h i t e}{\times \times \times \times x} 2 z = 4 \text{ } \leftarrow$ only one variable!
$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} \textcolor{m a \ge n t a}{z = 2}$

Substituting this value of 2 for $z$ into equation A gives:

$\textcolor{w h i t e}{\times \times \times \times \times} x + y \textcolor{m a \ge n t a}{+ 2} = 1$

This can be re-arranged to give $x$ as the subject:
$\textcolor{w h i t e}{\times \times \times x . \times} \textcolor{\lim e}{x = \left(- y - 1\right)}$

Now replace $\textcolor{\lim e}{x = \left(- y - 1\right)} \mathmr{and} \textcolor{m a \ge n t a}{z = 2} \text{ in } C$

$\textcolor{w h i t e}{\times \times \frac{\times}{.} x} 2 \textcolor{\lim e}{x} - 2 y \textcolor{m a \ge n t a}{- z} = - 16$..................C

$\textcolor{w h i t e}{\times} 2 \textcolor{\lim e}{\left(- y - 1\right)} - 2 y \textcolor{m a \ge n t a}{- 2} = - 16$..................C
$\textcolor{w h i t e}{x . \times} - 2 y - 2 - 2 y \textcolor{m a \ge n t a}{- 2} = - 16$..................C $\leftarrow$ only y !
$\textcolor{w h i t e}{x . \times \times \times \times . \times} - 4 y = - 16 + 4$

$\textcolor{w h i t e}{x . \times \times \times \times . \times} - 4 y = - 12$
$\textcolor{w h i t e}{x . \times \times \times \times . \times \times x} y = 3$

Now that we have a value for y, we can find x.

$\textcolor{\lim e}{x = - y - 1} = - \left(3\right) - 1$

$x = - 4$