How to find solution set in RR^3 of x-pi=5?

Using the method of substitution, obtain the solution set in ${\mathbb{R}}^{3}$ of the following: 1) $x - \pi = 5$ 2) $2 x - y + z = 1 , x - 2 y + z = 3 , y = \sqrt{2} - z$ 3) $x - y = 5 , x = 7 , 2 x - 3 y = 5$

Mar 16, 2017

1) $\left(x , y , z\right) = \left(\frac{57}{7} , 0 , 0\right)$
2) $\left(x , y , z\right) = \left(\frac{- 3 - \sqrt{2}}{4} , \frac{\sqrt{2} - 5}{4} , \frac{5 + 3 \sqrt{2}}{4}\right)$
3) $\left(x , y , z\right) = \left(7 , 2 , 0\right)$

Explanation:

1) $x - \pi = 5$
$x = 5 + \frac{22}{7} = \frac{57}{7}$
Therefore, $\left(x , y , z\right) = \left(\frac{57}{7} , 0 , 0\right)$

2) Solving using substitution method:
$2 x - y + z = 1$ (i)
$x - 2 y + z = 3$ (ii)
$y = \sqrt{2} - z$ (iii)

Solving (i) with substitution method we get:
$x = \frac{1}{2} + \frac{\sqrt{2}}{2} - z$

Putting value of x and y into (ii) we get:
$z = \frac{5 + 3 \sqrt{2}}{4}$

Putting value of z into (ii) we get:
$y = \frac{\sqrt{2} - 5}{4}$

Putting value of y and z into (i) we get:
$x = - \left(\frac{3 + \sqrt{2}}{4}\right)$

2) Solving using substitution method:
$x - y = 5$ (i)
$x = 7$ (ii)
$2 x - 3 y = 5$ (iii)

Putting the value of x into (i) we get:
$y = 2$

Therefore, $\left(x , y , z\right) = \left(7 , 2 , 0\right)$

Is there any explanation needed here to make the statement and solution more viable with the condition given (${\mathbb{R}}^{3}$)?