# How do you solve the simultaneous equations x+y+z=-2, 2x+5y+2z=-10, -x+6y-3z=-16 ?

Mar 25, 2017

$\left(x , y , z\right) = \left(- 2 , - 2 , 2\right)$

#### Explanation:

Given:

$\left\{\begin{matrix}x + y + z = - 2 \\ 2 x + 5 y + 2 z = - 10 \\ - x + 6 y - 3 z = - 16\end{matrix}\right.$

Subtracting twice the first equation from the second, we get:

$3 y = - 6$

Dividing both sides by $3$ we find:

$y = - 2$

Adding the first and third equation together, we get:

$7 y - 2 z = - 18$

Substituting $y = - 2$ into this equation, we get:

$- 14 - 2 z = - 18$

Add $14$ to both sides to get:

$- 2 z = - 4$

Divide both sides by $- 2$ to get:

$z = 2$

Then putting $y = - 2$ and $z = 2$ in the first equation, we find:

$x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = - 2$

Hence:

$x = - 2$

Mar 25, 2017

Use the 3 equations to write an Augmented Matrix and then perform elementary row operations until you obtain an identity matrix.

#### Explanation:

Write the augmented matrix:

[ (1,1,1,|,-2), (2,5,2,|,-10), (-1,6,-3,|,-16) ]

Perform elementary row operations.

$- 2 {R}_{1} + {R}_{2} \to {R}_{2}$

[ (1,1,1,|,-2), (0,3,0,|,-6), (-1,6,-3,|,-16) ]

${R}_{2} / 3$

[ (1,1,1,|,-2), (0,1,0,|,-2), (-1,6,-3,|,-16) ]

${R}_{1} + {R}_{3} \to {R}_{3}$

[ (1,1,1,|,-2), (0,1,0,|,-2), (0,7,-2,|,-18) ]

$- 7 {R}_{2} + {R}_{3} \to {R}_{3}$

[ (1,1,1,|,-2), (0,1,0,|,-2), (0,0,-2,|,-4) ]

${R}_{3} / - 2$

[ (1,1,1,|,-2), (0,1,0,|,-2), (0,0,1,|,2) ]

${R}_{1} - {R}_{2} \to {R}_{1}$

[ (1,0,1,|,0), (0,1,0,|,-2), (0,0,1,|,2) ]

${R}_{1} - {R}_{3} \to {R}_{1}$

[ (1,0,0,|,-2), (0,1,0,|,-2), (0,0,1,|,2) ]

We have obtained an identity matrix and the right column contains the solution set:

$x = - 2 , y = - 2 , \mathmr{and} z = 2$