# How do you solve the system of equations algebraically 7x+5y+z=0, -x+3y+2z=16, x-6y-z=-18?

Dec 4, 2017

Solution $x = - 2 , y = 2 , z = 4$ [Ans]

#### Explanation:

$7 x + 5 y + z = 0 \left(1\right) , - x + 3 y + 2 z = 16 \left(2\right)$ and

$x - 6 y - z = - 18 \left(3\right)$ from equation (1) we get $z = - 7 x - 5 y$

Putting $z = - 7 x - 5 y$ in equation (2) and (3) we get

$- x + 3 y - 14 x - 10 y = 16 \mathmr{and} - 15 x - 7 y = 16 \left(4\right)$ and

$x - 6 y + 7 x + 5 y = - 18 \mathmr{and} 8 x - y = - 18 \left(5\right)$ Multiplying

equation (5) by $7$ we get $56 x - 7 y = - 126 \left(6\right)$ Subtracting

equation (4) from equation (6) we get $71 x = - 142 \mathmr{and} x = - 2$

Putting $x = - 2$ in equation (1) and (2) we get

$5 y + z = 14 \left(7\right) \mathmr{and} 3 y + 2 z = 14 \left(8\right)$ Multiplying

equation (7) by $2$ we get $10 y + 2 z = 28 \left(9\right)$ Subtracting

equation (8) from equation (9) we get $7 y = 14 \mathmr{and} y = 2$

Putting $y = 2$ in equation (8) we get $3 \cdot 2 + 2 z = 14 \mathmr{and}$

$2 z = 8 \mathmr{and} z = 4$. Solution $x = - 2 , y = 2 , z = 4$ [Ans]

Dec 4, 2017

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

#### Explanation:

Adding third equation to first one,

$7 x + 5 y + z + x - 6 y - z = 0 - 18$ or $8 x - y = - 18$. $\left(1\right)$

Adding 2 times of third one to second one,

$- x + 3 y + 2 z + 2 \cdot \left(x - 6 y - z\right) = 16 - 2 \cdot 18$ or $x - 9 y = - 20$ $\left(2\right)$

Substracting 8 times of $\left(2\right)$ from $\left(1\right)$,

$8 x - y - 8 \cdot \left(x - 9 y\right) = - 18 - 8 \cdot \left(- 20\right)$

$71 y = 142$, so $y = 2$

Hence,

$x - 9 \cdot 2 = - 20$ or $x - 18 = - 20$, so $x = - 2$

Thus,

$7 \cdot \left(- 2\right) + 5 \cdot 2 + z = 0$ or $z - 4 = 0$, so $z = 4$