Tigonometry help!!!?

Question

x-2y+5z=2
4x-z=0
I need help..the answer in the back on the book
is (2a,21a-1,8a)
can you please explain to me how to do it

Answer 1

i dont think this is tigonometry, not even trigonometry, this is a problem in simultaneous equations...

x-2y+5z=2
4x-z=0

from the second equation: z= 4x, substitute it in the first equation:
x-2y + 5(4x) =2 (now we have only 2 variables,
x-2y +20x =2
21x -2y =2
2y = 21x -2
y=(21x-2)/2

and there is no way of reducing the number of variables any further, so we set a=x, then:
x=a
y= ( 21a-2) /2
z=4a
or in vector notation:
(a,(21a-2)/2,4a)

now this is equivalent to the solution given in the back of the book . .

Answer 2

4x - z = 0 => z = 4x, so substitute this into the first equation:

x - 2y + 5(4x) = 2
=> 21x - 2y = 2
=> 21x = 2y + 2

Since we only have two equations and three variables, we will have infinitely many answers. Now 21 and 2 have lowest common multiple 42, so write
21x = 42a = 2y + 2
Then x = 2a, y = 21a - 1 and z = 4x = 8a. So the solution is
(2a, 21a - 1, 8a) for any a in R.

Note that the same solution can be expressed differently. For instance, we could have written
21x - 2 = 42b = 2y
Then x = 2b + 2/21, y = 21b, z = 4x = 8b + 8/21, so we'd get
(2b + 2/21, 21b, 8b + 8/21) for any b in R.

This is the same solution; you should be able to see that if you set b = a + 1/21 these are the same.