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

Question

Using the method of substitution, obtain the solution set in $RR^3$ of the following:

1) $x-pi=5$
2) $2x-y+z=1, x-2y+z=3, y=sqrt2-z$
3) $x-y=5, x=7, 2x-3y=5$

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Answer 1 · 2017-03-16T21:11:10

1) $(x,y,z)=(57/7, 0, 0)$
2) $(x, y, z) = ((-3-sqrt2)/4, (sqrt2-5)/4, (5+3sqrt2)/4)$
3) $(x, y, z) = (7,2,0)$

1) $x-pi=5$
$x=5+22/7 = 57/7$
Therefore, $(x,y,z)=(57/7, 0, 0)$

2) Solving using substitution method:
$2x-y+z=1$ (i)
$x-2y+z=3$ (ii)
$y=sqrt2-z$ (iii)

Solving (i) with substitution method we get:
$x=1/2 + sqrt2/2 - z$

Putting value of x and y into (ii) we get:
$z= (5+3sqrt2)/4$

Putting value of z into (ii) we get:
$y = (sqrt2-5)/4$

Putting value of y and z into (i) we get:
$x= -((3+sqrt2)/4)$

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

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

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

Is there any explanation needed here to make the statement and solution more viable with the condition given ( $RR^3$ )?